course PHY 231 Something weird was happening with my computer at some points of this assignment and I think a few of the answers I put are not even posted. ܤr}剫 ܊D}assignment #027
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01:08:43 Query intro probs set 7, 1-7 Knowing the 9.8 m/s^2 gravitational field strength of the Earth's field at the surface of the Earth, and knowing the radius of the Earth, how do we find the gravitational field strength at a given distance 'above' the surface of the Earth?
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RESPONSE --> Field strenght is found by dividing acceleration by the distance between centers of two masses
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01:17:00 ** You have an inverse square force. Square the ratio of Earth radius to orbital radius and multiply by 9.8 m/s^2: Field strength=(Re/r)^2*9.8m/s^2 **
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RESPONSE --> I was not sure how to find field strength.
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02:38:18 If we double our distance from the center of the Earth, what happens to the gravitational field strength we experience?
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RESPONSE --> Force is inversely proportional to the square of distance. If the distance is doubled, then the force would be a fourth of its original strength.,
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02:38:24 ** We have an inverse square force so if r2 = 2 * r1 the ratio of the gravitational field will be g2 / g1 = (1 / r2^2) / (1 / r1^2) = r1^2 / r2^2 = (r1 / r2)^2 = (r1 / (2 * r1))^2 = r1^2 / 4 r1^2 = 1/4. In a nutshell double the radius gives us 1 / 2^2 = 1/4 the gravitational field. **
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RESPONSE --> ok
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02:46:26 How do we approximate the energy required to move a given mass from the surface of the Earth to a given height 'above' the Earth, where the field strength at the given height differ significantly from that at the surface?
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RESPONSE --> In the force experiment, the r is a distance between centers not simply a radius of the planet or object.
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02:49:25 STUDENT SOLUTION AND INSTRUCTOR RESPONSE: mass*[(Re + distance)/Re]^2=force Force*distance=KE INSTRUCTOR RESPONSE: The first approximation would be to average the force at the surface and the force at the maximum altitude, then multiply by the distance. The result would give you the work necessary to 'raise' the object against a conservative force, which would be equal to the change in PE. ADDENDUM FOR UNIVERSITY PHYSICS STUDENTS ONLY:The exact work is obtained by integrating the force with respect to position. You can integrate either G M m / r^2 or g * (RE / r)^2 from r = RE to rMax. **
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RESPONSE --> I misinterpreted the question. I read the chapter but am still getting accustomed to the force equation involvin Re + distance / re^2.
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02:49:29 Query class notes #24
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02:51:46 Describe the paths of various particles 'shot' parallel to the surface of the Earth from the top of a very high tower, starting with a very small velocity and gradually increasing to a velocity sufficient to completely escape the gravitational field of the Earth.
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RESPONSE --> Particles shot parallel to the earth would have a centripetal motion that is the same velocity as the rotation of the earth. The higher it is, the farther away it is, which means that the force is not as strong.
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02:55:15 GOOD STUDENT ANSWER: Each particle sets out to follow an orbit around the center of mass of the earth. But for particles shot at slower speeds, this path is interupted by the surface of the eath and simply stops there. The faster it is shot, the further x distance becomes before the particle lands. However, if it given a great enough velocity, it will fall around the curviture of the earth. If is shot even faster than that, it will follow an eliptical oribit with varying speeds and distances from center of earth. GOOD STUDENT ANSWER: With a very low velocity the projectile will not travlel as far. It will fall to earth in a nearly parabolic fashion since it gains vertical velocity as it travels horizontally at a steady pace. If the projectile is fired at a very strong velocity it will leave the earths vacinity but will still be pulled by the forces acting on it from the earths center. This will cause it to go only so far at which point it has slowed down considerabley, since it has lost most of its kinetic energy. It turns and begins to gain energy as it approaches the earths area, using the potential energy it gained on the trip out. (Causing it to speed up). The path that this projectile will take will be eliptical, and it will continue to loop around the earth. If the projectile is fired at the correct velocity to form a circular orbit, it will also fall at a parabolic fashion, although the earth's surface will also be descending at the same rate so that the object will appear to be 'not falling'. It is falling but at the same rate the earth is 'falling' under it. It will circle the earth until something causes it to stop. INSTRUCTOR RESPONSE: The path of the projectile will always be an ellipse with the center of the Earth at one focus. For low velocities and low altitude this path is very nearly parabolic before being interrupted by the surface of the Earth. One of these ellipses is a perfect circle and gives us the circular orbit we use frequently in this section. **
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02:55:55 How many of the velocities in the preceding question would result in a perfectly circular orbit about the Earth?
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RESPONSE --> There would be only 1. It has to be at a far enough distance from the earth to travel at the rotation speed of the earth.
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02:56:06 ** For a given distance from the center of the Earth, there is only one velocity for which centripetal acceleration is equal to gravitational acceleration, so there is only one possible velocity for a circular orbit of given orbital radius. The orbital radius is determined by the height of the 'tower', so for a given tower there is only one velocity which will achieve a circular orbit. **
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RESPONSE --> ok
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02:56:30 Is it necessary in order to achieve a circular orbit to start the object out in a direction parallel to the surface of the Earth?
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RESPONSE --> Yes it must be tangent to the circular path of the earth.
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02:56:45 ** If you have just one 'shot' then you must start out parallel to the surface of the Earth. The reason is that any circle about the center must be perpendicular at every point to a radial line--a line drawn from the center to the circle. Any radial line will intercept the surface of the Earth and must be perpendicular to it, and the circular orbit must also be perpendicular to this line. Therefore the orbit and the surface are perpendicular to the same line and are therefore parallel. **
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02:56:47 Principles of Physics and General College Physics Problem 5.2: A jet traveling at 525 m/s moves in an arc of radius 6.00 km. What is the acceleration of the jet?
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02:56:49 The jet will have centripetal acceleration a_cent = v^2 / r, where v is its speed and r the radius of the circle on which it is traveling. In this case we have v = 525 m/s and r = 6.00 km = 6000 meters. The centripetal acceleration is therefore a_cent = v^2 / r = (525 m/s)^2 / (6000 m) = 45 m/s^2, approx.. One 'g' is 9.8 m/s^2, so this is about (45 m/s^2) / (9.8 m/s^2) = 4.6 'g's'.
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02:57:16 Univ. Why is it that the center of mass doesn't move?
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RESPONSE --> The center of mass will move but at such a small distance that it is not noticeable.
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vzwxj^i assignment #028 骶͌Άrzvc֡ Physics I 08-06-2008
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03:52:33 Query class notes #26 Explain how we use proportionality along with the radius rE of the Earth to determine the gravitational acceleration at distance r from the center of the Earth to obtain an expression for the gravitational acceleration at this distance. Explain how we use this expression and the fact that centripetal forces is equal to v^2 / r to obtain the velocity of a satellite in circular orbit.
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RESPONSE --> Velocity in the orbit's acceleration must be placed a correct distance from the center of the earth in order to move at the speed of the earth's rotation.
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03:52:57 ** The proportionality is accel = k r^2. When r = rE, accel = 9.8 m/s^2 so 9.8 m/s^2 = k * rE^2. Thus k = 9.8 m/s^2 / rE^2, and the proportionality can now be written accel = [ 9.8 m/s^2 / (rE)^2 ] * r^2. Rearranging this gives us accel = 9.8 m/s^2 ( r / rE ) ^2. **
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RESPONSE --> I am still having some difficulty witht this.
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03:52:59 Principles of Physics and Gen Phy problem 5.30 accel of gravity on Moon where radius is 1.74 * 10^6 m and mass 7.35 * 10^22 kg.
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03:53:01 ** The acceleration due to gravity on the Moon is found using the equation g' = G (Mass of Moon)/ radius of moon ^2 g' = (6.67 x 10^-11 N*m^2/kg^2)(7.35 X 10^22 kg) / (1.74 X 10^6 m) = 1.619 m/s^2 **
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03:53:03 Query gen phy problem 5.40 force due to planets (Mv, Mj, Ms, are .815, 318, 95.1 Me; orb radii 108, 150, 778, 1430 million km). What is the total force on Earth due to the planets, assuming perfect alignment?
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03:53:04 ** Using F = G m1 m2 / r^2 we get Force due to Venus: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (.815 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 1.08 * 10^11 m)^2 = 1.1 * 10^18 N, approx. Force due to Jupiter: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (318 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 7.78 * 10^11 m)^2 = 1.9 * 10^18 N, approx. Force due to Saturn: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (95.7 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 1.43 * 10^11 m)^2 = 1.4 * 10^17 N, approx. Venus being 'inside' the Earth's orbit pulls in the direction of the Sun while Jupiter and Saturn pull in the opposite direction so the net force is -1.1 * 10^18 N + 1.9 * 10^18 N + 1.4 * 10^17 N = .9 * 10^18 N = 9 * 10^17 N, approx.. **
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03:53:30 Univ. 12.50 (12.44 10th edition). 25 kg, 100 kg initially 40 m apart, deep space. Both objects have identical radii of .20 m. When 20 m apart what is the speed of each (relative to the initial common speed, we presume), and what is the velocity relative to one another? Where do they collide? Why does position of center of mass not change?
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03:54:29 The force would be F = (6.67 * 10^-11 * 25 * 100) / 20^2 F = 4.17 * 10^-10 a1 = 4.17 * 10^-10 / 25 a1 = 1.67 * 10^-11 m/s/s a2 = 4.17 * 10^-10 / 100 a2 = 4.17 * 10^-12 m/s/s The position of center of mass doesn't change because the two spheres are the same size. ** At separation r the force is F = G m1 m2 / r^2. For any small increment `dr of change in separation the approximate work done by the gravitational force is F `dr = G m1 m2 / r^2 * `dr. We take the sum of such contributions, between the given separations, to form an approximation to the total work done by the gravitational force. We then take the limit as `dr -> 0 and obtain the integral of G m1 m2 / r^2 with respect to r from separation r1 to separation r2. An antiderivative is - G m1 m2 / r; evaluating between the two separations we get - G m1 m2 / r1 - (-G m1 m2 / r2) = G m1 m2 ( 1/r2 - 1 / r1). This expression is evaluated at r1 = 40 m and r2 = 20 m to get the change G m1 m2 ( 1/(20 m) - 1 / (40 m) ) in KE. I get around 1.49 * 10^-9 Joules but it isn't guaranteed so you should verify that carefully. Assuming a reference frame initially at rest with respect to the masses the intial momentum is zero. If the velocities at the 20 m separation are v1 and v2 we know that m1 v1 + m2 v2 = 0, so that v2 = -(m1 / m2) * v1. The total KE, which we found above, is .5 m1 v1^2 + .5 m2 v2^2. Substituting v2 = - (m1 / m2) v1 and setting equal to the KE we can find v1; from this we easily find v2. You might get something like 4.1 * 10^-6 m/s for the velocity of the 100 kg mass; this number is again not guaranteed so verify it yourself. The position of the center of mass does not change because there is no external force acting on the 2-mass system. The center of mass is at position r with respect to m1 (take m1 to be the 25 kg object) such that m1 r - m2 (40 meters -r) = 0; substituting m1 and m2 you get 25 r - 100 (40 meters - r ) = 0. I believe you get r = 4 / 5 * 40 meters = 32 m, approx., from the 25 kg mass, which would be 8 meters from the 100 kg mass.
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03:54:31 Query gen phy problem 5.50 24 m diam wheel, rot period 12.5 s, fractional change in apparent weight at top and at bottom. What is the fractional change in apparent weight at the top and that the bottom of the Ferris wheel?
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03:54:33 ** Centripetal acceleration is a = v^2 / r. For a point on the rim of the wheel, v = dist in 1 rev / time for 1 rev = `pi * 24 m / (12.5 sec) = 1.9 m/s, approx. Thus v^2 / r = (`pi * 1.9 m/s)^2 / 12 m = 3 m/s^2, approx. At the top the only accel is the centripetal, and it is acting toward the center, therefore downward. The forces acting on any mass at the top are the gravitational force and the force exerted by the wheel on the mass. At the top of the wheel the latter force is the apparent weight. Thus grav force + apparent weight = centripetal force - m * 9.8 m/s^2 + wtApparent = m * (-3 m/s^2 ) wtApparent = m (-3 m/s^2) + m ( 9.8 m/s^2) = m (6.8 m/s^2). A similar analysis at the bottom, where the centripetal force will be toward the center, therefore upward, gives us - m * 9.8 m/s^2 + wtApparent = m * (+3 m/s^2 ) wtApparent = m (+3 m/s^2) + m ( 9.8 m/s^2) = m (12.8 m/s^2). The ratio of weights is thus 12.8 / 6.8, approx. ** A more elegant solution obtains the centripetal force for this situation symbolically: Centripetal accel is v^2 / r. Since for a point on the rim we have v = `pi * diam / period = `pi * 2 * r / period, we obtain aCent = v^2 / r = [ 4 `pi^2 r^2 / period^2 ] / r = 4 `pi^2 r / period^2. For the present case r = 12 meters and period is 12.5 sec so aCent = 4 `pi^2 * 12 m / (12.5 sec)^2 = 3 m/s^2, approx. This gives the same results as before. **
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03:55:06 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> That one problem for University Physics had a lot of calculus and was a little confusing. I was not sure what to do.
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G~qėjƕ՝ assignment #029 骶͌Άrzvc֡ Physics I 08-06-2008
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09:56:26 Query class notes #28. Explain how we can calculate the average angular velocity and the angular acceleration of an object which rotates from rest through a given angle in a given time interval, assuming constant angular acceleration.
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RESPONSE --> Angular velocity is found by dividing velocity by radians. Acceleration would be found by multiplying angular velocity and the radians.
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09:57:23 **This situation is strictly analogous to the one you encountered early in the course. As before acceleration is change in velocity / change in clock time. However now it's angular acceleration. We have angular acceleration = change in angular velocity / change in clock time. The average angular velocity is change in angular position / change in clock time. This question assumes you know the angle through which the object rotates, which is its change in angular position, as well as the change in clock time. So you can calculate the average angular velocity. If angular accel is uniform and initial angular velocity is zero then the final angular velocity is double the average angular velocity. In this case the change in angular velocity is equal to the final angular velocity, which is double the average angular velocity. From this information you can calculate angular acceleration. **
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RESPONSE --> ok
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09:57:25 Principles of Physics and General College Physics Problem 7.46: Center of mass of system 1.00 kg at .50 m to left of 1.50 kg, which is in turn .25 m to left of 1.10 kg.
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09:57:27 Using the position of the 1.00 kg mass as the x = 0 position, the other two objects are respectively at x = .50 m and x = .75 m. The total moment of the three masses about the x = 0 position is 1.00 kg * (0 m) + 1.50 kg * (.50 m) + 1.10 kg * (.75 m) = 1.58 kg m. The total mass is 1.00 kg + 1.50 kg + 1.10 kg = 3.60 kg, so the center of mass is at position x_cm = 1.58 kg m / (3.60 kg) = .44 meters, placing it a bit to the left of the 1.50 kg mass.
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09:57:29 Query problem 7.50 3 cubes sides L0, 2L0 and 3L0; center of mass.
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09:57:31
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09:57:33 ** The mass of the second will be 2^3 = 8 times as great as the first. It takes 8 1-unit cubes to make a 2-unit cube. The mass of the third will be 3^3 = 27 times as great as the first. It takes 27 1-unit cubes to make a 3-unit cube. In the x direction the distance from left edge to center of first cube is 1/2 L0 (the center of the first cube). In the y direction the distance is from lower edge to center of the first cube is 1/2 L0 (the center of the first cube). In the x direction the distance from left edge to center of the second cube is L0 + L0 (the L0 across the first cube, another L0 to the center of the second), or 2 L0. In the y direction the distance from lower edge to center of the second cube is L0 (the center of the second cube). In the x direction the distance from left edge to center of the third cube is L0 + 2 L0 + 3/2 L0 (the L0 across the first cube, another 2 L0 across the second and half of 3L0 to the center of the third), or 9/2 L0. In the x direction the distance from lower edge to center of the first cube is 3/2 L0 (the center of the third cube). Moments about left edge and lower edge of first cube: If m1 is the mass of the first cube then in the x direction you have total moment m1 * L0/2 + 8 m1 * (2 L0) + 27 M1 * (9/2 L0) = 276 m1 L0 / 2 = 138 m1 L0. The total mass is m1 + 8 m1 + 27 m1 = 36 m1 so the center of mass is at center of mass in x direction: 138 m1 L0 / (36 m1) = 3.83 L0. In the y direction the moment is m1 * L0/2 + 8 m1 * (L0) + 27 m1 * ( 3/2 L0) = 45 m1 L0 so the center of mass is at center of mass in y direction: 45 m1 L0 / (36 m1) = 1.25 L0. **
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09:57:36 What is the mass of the second cube as a multiple of the mass of the first?
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09:57:37 ** 3 dimensions: the mass will be 2^3 = 8 times as great. It takes 8 1-unit cubes to make a 2-unit cube. **
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09:57:39 What is the mass of the third cube as a multiple of the mass of the first?
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09:57:40 ** The mass of the third cube is 3^3 = 27 times the mass of the first. **
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09:57:42 How far from the outside edge of the first cube is its center of mass?
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09:57:43 ** In the x direction the distance is 1/2 L0 (the center of the first cube). In the y direction the distance is also 1/2 L0 (the center of the first cube). **
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09:57:44 How far from the outside edge of the first cube is the center of mass of the second cube?
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09:57:46 ** In the x direction the distance is L0 + L0 (the L0 across the first cube, another L0 to the center of the second), or 2 L0. In the x direction the distance is L0 (the center of the second cube). **
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09:57:47 How far from the outside edge of the first cube is the center of mass of the third cube?
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09:57:48 ** In the x direction the distance is L0 + 2 L0 + 3/2 L0 (the L0 across the first cube, another 2 L0 across the second and half of 3L0 to the center of the third), or 9/2 L0. In the x direction the distance is 3/2 L0 (the center of the third cube). **
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09:57:49 How do you use these positions and the masses of the cubes to determine the position of the center of mass of the system?
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09:57:51 ** In the x direction you have moment m1 * L0/2 + 8 m1 * (2 L0) + 27 M1 * (9/2 L0) = 276 m1 L0 / 2 = 138 m1 L0. The total mass is m1 + 8 m1 + 27 m1 = 36 m1 so the center of mass is at 138 m1 L0 / (36 m1) = 3.83 L0. In the y direction the moment is m1 * L0/2 + 8 m1 * (L0) + 27 m1 * ( 3/2 L0) = 45 m1 L0 so the center of mass is at 45 m1 L0 / (36 m1) = 1.25 L0. **
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09:58:31 Univ. 8.94 (8.82 10th edition). 45 kg woman 60 kg canoe walk starting 1 m from left end to 1 m from right end, moving 3 meters closer to the right end. How far does the canoe move? Water resistance negligible.
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RESPONSE --> If the canoe is 3 meters and it is said to move 1 meter, then it moves 1 meter.
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09:59:16 ** Since water resistance is negligible the net force acting on the system is zero. Since the system is initially stationary the center of mass of the system is at rest; since zero net force acts on the system this will continue to be the case. Assuming that the center of mass of the canoe is at the center of the canoe, then when the woman is 1 m from the left end the center of mass of the system lies at distance c.m.1 = (1 m * 45 kg + 2.5 m * 60 kg) / (45 kg + 60 kg) = 195 kg m / (105 kg) = 1.85 m from the left end of the canoe. A similar analysis shows that when the woman is 1 m from the right end of the canoe, then since she is 4 m from the left end the center of mass lies at c.m.2 = (4 m * 45 kg + 2.5 m * 60 kg) / (45 kg + 60 kg) = 310 kg m / (105 kg) = 2.97 m. The center of mass therefore changes its position with respect to the left end of the canoe by about 1.1 meters toward the right end of the canoe. Since the center of mass itself doesn't move the canoe must move 1.1 meters toward the left end, i.e., backwards. Note that since the woman moves 3 m forward with respect to the canoe and the canoe moves 1.3 m backwards the woman actually moves 1.7 m forward. The sum -1.3 m * 60 kg + 1.7 m * 45 kg is zero, to within roundoff error. This is as it should be since this sum represents the sum of the changes in the centers of mass of the canoe and the woman, which is the net change in the position of center of mass. **
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Bڕst assignment #030 骶͌Άrzvc֡ Physics I 08-06-2008
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10:09:58 introductory set 8. If we know the constant moment of inertia of a rotating object and the constant net torque on the object, then how do we determine the angle through which it will rotate, starting from rest, in a given time interval?
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RESPONSE --> If we multiply net torque and constant inertia, we can find the angle through which it will rotate.
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10:10:35 ** tau stands for torque and I stands for the moment of inertia. These quantities are analogous to force and mass. Just as F = m a, we have tau = I * alpha; i.e., torque = moment of inertia * angular acceleration. If we know the moment of inertia and the torque we can find the angular acceleration. If we multiply angular acceleration by time interval we get change in angular velocity. We add the change in angular velocity to the initial angular velocity to get the final angular velocity. In this case initial angular velocity is zero so final angular velocity is equal to the change in angular velocity. If we average initial velocity with final velocity then, if angular accel is constant, we get average angular velocity. In this case angular accel is constant and init vel is zero, so ave angular vel is half of final angular vel. When we multiply the average angular velocity by the time interval we get the angular displacement, i.e., the angle through which the object moves. **
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RESPONSE --> ok
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10:11:45 If we know the initial angular velocity of a rotating object, and if we know its angular velocity after a given time, then if we also know the net constant torque accelerating the object, how would we find its constant moment of inertia?
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RESPONSE --> Torque is found by multiplying inertia and average angular velocity. Find average velocity by summing up initial and final velocities then dividing by 2. Divide torque by average velocity and you have constant moment of inertai
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10:11:53 ** From init and final angular vel you find change in angular vel (`d`omega = `omegaf - `omega0). You can from this and the given time interval find Angular accel = change in angular vel / change in clock time. Then from the known torque and angular acceleration we find moment of intertia. tau = I * alpha so I = tau / alpha. **
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10:13:22 How do we find the moment of inertia of a concentric configuration of 3 uniform hoops, given the mass and radius of each?
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RESPONSE --> Find the mass and radius of each multiply the mass and square of the radius. Then take the sum of each hoop to find the inertia.
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10:13:27 ** Moment of inertia of a hoop is M R^2. We would get a total of M1 R1^2 + M2 R2^2 + M3 R3^2. **
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10:13:43 How do we find the moment of inertia a light beam to which are attached 3 masses, each of known mass and lying at a known distance from the axis of rotation?
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RESPONSE --> Same equation
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10:13:49 ** Moment of inertia of a mass r at distance r is m r^2. We would get a total of m1 r1^2 + m2 r2^2 + m3 r3^2. Note the similarity to the expression for the hoops. **
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10:13:52 Principles of Physics and General College Physics problem 8.4. Angular acceleration of blender blades slowing to rest from 6500 rmp in 3.0 seconds.
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10:13:53 The change in angular velocity from 6500 rpm to rest is -6500 rpm. This change occurs in 3.0 sec, so the average rate of change of angular velocity with respect to clock time is ave rate = change in angular velocity / change in clock time = -6500 rpm / (3.0 sec) = -2200 rpm / sec. This reasoning should be very clear from the definition of average rate of change. Symbolically the angular velocity changes from omega_0 = 6500 rpm to omega_f = 0, so the change in velocity is `dOmega = omega_f - omega_0 = 0 - 6500 rpm = -6500 rpm. This change occurs in time interval `dt = 3.0 sec. The average rate of change of angular velocity with respect to clock time is therefore ave rate = change in angular vel / change in clock time = `dOmega / `dt = (omega_f - omega_0) / `dt = (0 - 6500 rpm) / (3 sec) = -2200 rpm / sec. The unit rpm / sec is a perfectly valid unit for rate of change of angular velocity, however it is not the standard unit. The standard unit for angular velocity is the radian / second, and to put the answer into standard units we must express the change in angular velocity in radians / second. Since 1 revolution corresponds to an angular displacement of 2 pi radians, and since 60 seconds = 1 minute, it follows that 1 rpm = 1 revolution / minute = 2 pi radians / 60 second = pi/30 rad / sec. Thus our conversion factor between rpm and rad/sec is (pi/30 rad / sec) / (rpm) and our 2200 rpm / sec becomes angular acceleration = 2200 rpm / sec * (pi/30 rad / sec) / rpm = (2200 pi / 30) rad / sec^2 = 73 pi rad / sec^2, or about 210 rad / sec^2.
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10:13:55 Principles of Physics and General College Physics problem 8.16. Automobile engine slows from 4500 rpm to 1200 rpm in 2.5 sec. Assuming constant angular acceleration, what is the angular acceleration and how how many revolutions does the engine make in this time?
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10:13:56 The change in angular velocity is -3300 rpm, which occurs in 2.5 sec. So the angular acceleration is angular accel = rate of change of angular vel with respect to clock time = -3300 rpm / (2.5 sec) = 1300 rpm / sec. Converting to radians / sec this is about angular accel = -1300 rpm / sec ( pi / 30 rad/sec) / rpm = 43 pi rad/sec^2, approx.. Since angular acceleration is assumed constant, a graph of angular velocity vs. clock time will be linear so that the average angular velocity with be the average of the initial and final angular velocities: ave angular velocity = (4500 rpm + 1200 rpm) / 2 = 2750 rpm, so that the angular displacement is angular displacement = ave angular velocity * time interval = 2750 rpm * 2.5 sec = 6900 revolutions, approximately. In symbols, using the equations of uniformly accelerated motion, we could use the first equation `dTheta = (omega_0 + omega_f) / 2 * `dt = (4500 rpm + 1200 rpm) / 2 * (2.5 sec) = 6900 revolutions and the second equation omega_f = omega_0 + alpha * `dt, which is solved for alpha to get alpha = (omega_f - omega_0) / `dt = (4500 rpm - 1200 rpm) / (2.5 sec) = 1300 rpm / sec, which as before can be converted to about 43 pi rad/sec^2, or about 130 rad/sec^2. The angular displacement of 6900 revolutions can also be expressed in radians as 6900 rev = 6900 rev (2 pi rad / rev) = 13800 pi rad, or about 42,000 radians.
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10:14:26 Univ. 9.72 (64 in 10th edition). motor 3450 rpm, saw shaft 1/2 diam of motor shaft, blade diam .208 m, block shot off at speed of rim. How fast and what is centrip accel of pt on rim?
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10:14:28 ** The angular velocity of the shaft driving the blade is double that of the motor, or 3450 rpm * 2 = 7900 rpm. Angular velocity is 7900 rpm = 7900 * 2 pi rad / 60 sec = 230 pi rad / sec. At a distance of .208 m from the axis of rotation the velocity will be .208 m * 230 pi rad / sec = 150 m/s, approx.. The angular acceleration at the .208 m distance is aCent = v^2 / r = (150 m/s)^2 / (.208 m) = 108,000 m/s^2, approx.. The electrostatic force of attraction between sawdust and blade is nowhere near sufficient to provide this much acceleration. **
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ϨOS assignment #031 骶͌Άrzvc֡ Physics I 08-06-2008
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10:51:00 experiment to be viewed. What is the relationship between the angular velocity of the axle around which the string is wound and that of the large disk?
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RESPONSE --> They are proportional, Even though the larger disk covers more area, its velocity is greater than the small disk.
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10:51:21 GOOD STUDENT RESPONSE The angular velocity of the axle and the angular velocity of the disk on the axle would be the same. However, the velocity would be different because they are of different distances from the center. In general, the axle will be moving at a slower speed(velocity) than a point on the outside of the disk. I am not sure if this is what you are asking. ** The speed of the falling object is the same as the speed of a point on the rim of the axle. The angular velocity of the axle is equal to the speed of a point on its rim divided by its radius: omega = v / r. The disk rotates with the axle so it has the same angular velocity. **
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10:51:47 If the falling weight accelerates uniformly, does it follow that the rotating disk has a uniform angular acceleration?
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RESPONSE --> Uniform acceleration would mean that every component of the disk accelerates at the same rate.
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10:51:55 GOOD STUDENT RESPONSE yes, because the angle of acceleration is proportional to the velocity of the disk with the radius(which is constant) as the constant of proportionality. And the velocity of the disk will be the same as the velocity of the falling weight which is dependent on the acceleration of the weight. ** If v changes at a uniform rate then since r is uniform, omega = v / r changes at a uniform rate. **
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10:51:57 Principles of Physics and General College Physics Problem 8.28: Moment of inertia of bicycle wheel 66.7 cm diameter, mass 1.25 kg at rim and tire.
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10:51:58 The mass of the rim and tire is all located at about the same distance from the axis of rotation, so the rim and tire contribute m * r^2 to the total moment of inertia, where m is the mass and r the distance from the axis of rotation of the rim and tire. The distance r is half the diameter, or 1/2 * 66.7 cm = 33.4 cm = .334 m, and the mass is given as 1.25 kg, so the moment of inertia of rim and tire is I = m r^2 = 1.25 kg * (.334 m)^2 = 1.4 kg m^2.
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10:52:00 Why can the mass of the hub be ignored?
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10:52:01 The radius of the hub is less than 1/5 the radius of the tire; because its moment of inertia is m r^2, where r is its 'average' distance from the axis of rotation, its r^2 will be less than 1/25 as great as for the rim and tire. Even if the mass of the hub is comparable to that of the rim and tire, the 1/25 factor will make its contribution to the moment of inertia pretty much negligible.
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10:52:02 gen Problem 8.38 arm, 3.6 kg ball accel at 7 m/s^2, triceps attachment 2.5 cm below pivot, ball 30 cm above pivot. Give your solution to the problem.
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10:52:03 ** The moment of inertia of a 3.6 kg ball at a point 30 cm from the axis of rotation is I = m r^2 = 3.6 kg * (.30 m)^2 = .324 kg m^2. At a 30 cm distance from axis of rotation the 7 m/s^2 acceleration becomes an angular acceleration of alpha = a / r = 7 m/s^2 / (.3 m) = 23.3 rad/s^2. The necessary torque is therefore tau = I * alpha = .324 kg m^2 * 23.3 rad/s^2 = 7.6 m N, approx.. The muscle exerts its force at a point x = 2.5 cm from the axis of rotation and perpendicular to that axis so we have F = tau / x = 7.6 m N / (.025 m) = 304 N. **
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10:52:05 Univ. 10.52 (10.44 10th edition). 55 kg wheel .52 m diam ax pressed into wheel 160 N normal force mu =.60. 6.5 m N friction torque; crank handle .5 m long; bring to 120 rev/min in 9 sec; torque required? Force to maintain 120 rev/min? How long to coast to rest if ax removed?
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10:52:06 ** The system is brought from rest to a final angular velocity of 120 rev/min * 1min/60 sec * 2`pi/1 rev = 12.6 rad/s. The angular acceleration is therefore alpha = change in omega / change in t = 12.6 rad/s / (9 sec) = 1.4 rad/s^2, approx.. The wheel has moment of inertia I = .5 m r^2 = .5 * 55 kg * (.52 m)^2 = 7.5 kg m^2, approx.. To achieve the necessary angular acceleration we have tauNet = I * alpha = 7.5 kg m^2 * 1.4 rad/s^2 = 10.5 m N. The frictional force between ax and wheel is .60 * 160 N = 96 N at the rim of the wheel, resulting in torque tauFrictAx = -96 N * .52 m = -50 m N. The frictional torque of the wheel is in the direction opposite motion and is therefore tauFrict = -6.5 m N. The net torque is the sum of the torques exerted by the crank and friction: tauNet = tauFrictAx + tauFrict + tauCrank so that the torque necessary from the crank is tauCrank = tauNet - tauFrict - tauCrank = 10.5 m N - (-50 m N) - (-6.5 m N) = 67 m N. The crank is .5 m long; the force necessary to achive the 60.5 m N torque is therefore F = tau / x = 67 m N / (.5 m) = 134 N. If the ax is removed then the net torque is just the frictional torque -6.5 m N so angular acceleration is alpha = -6.5 m N / (7.5 kg m^2) = -.84 rad/s^2 approx. Starting at 120 rpm = 12.6 rad/s the time to come to rest will be `dt = `dOmega / alpha = -12.6 rad/s / (-.84 rad/s^2) = 14.5 sec, approx.. **
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{e˚}ďЉ assignment #032 骶͌Άrzvc֡ Physics I 08-06-2008
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11:00:14 Query experiment to be viewed. What part or parts of the system experiences a potential energy decrease? What part or parts of the system experience(s) a kinetic energy increase?
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RESPONSE --> I cannot view the experiment
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11:00:15 ** The mass on the string descends and loses PE. The wheel and the descending mass both increase in KE, as do the other less massive parts of the system (e.g., the string) and slower-moving parts (e.g., the axel, which rotates at the same rate as the wheel but which due to its much smaller radius does not move nearly as fast as most of the wheel). **
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11:00:17 What part or parts of the system experience(s) an increase in angular kinetic energy?
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11:00:18 ** The wheel, the bolts, the axle, and anything else that's rotating experiences an increase in angular KE. **
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11:00:20 What part or parts of the system experience(s) an increasing translational kinetic energy?
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11:00:21 ** Only the descending mass experiences an increase in translational KE. **
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11:00:23 Does any of the bolts attached to the Styrofoam wheel gain more kinetic energy than some other bolt? Explain.
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11:00:24 ** The bolts toward the outside of the wheel are moving at a greater velocity relative to some fixed point, so their kinetic energy is greater since k = 1/2 m v^2 **
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11:00:26 What is the moment of inertia of the Styrofoam wheel and its bolts?
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11:00:27 ** The moment of inertia for the center of its mass=its radias times angular velocity. Moment of inertia of a bolt is m r^2, where m is the mass and r is the distance from the center of mass. The moment of inertia of the styrofoam wheel is .5 M R^2, where M is its mass and R its radius. The wheel with its bolts has a moment of inertia which is equal to the sum of all these components. **
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11:01:06 How do we determine the angular kinetic energy of of wheel by measuring the motion of the falling mass?
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RESPONSE --> Kinetic Energy is found by multiplying an object's angular speed by moment of inertia.
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11:01:20 ** STUDENT ANSWER AND INSTRUCTOR CRITIQUE: The mass falls at a constant acceleration, so the wheel also turns this fast. INSTRUCTOR CRITIQUE: We don't use the acceleration to find the angular KE, we use the velocity. The acceleration, if known, can be used to find the velocity. However in this case what we are really interested in is the final velocity of the falling mass, which is equal to the velocity of the part of the wheel around which it is wound. If we divide the velocity of this part of the wheel by the its radius we get the angular velocity of the wheel. **
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11:01:22 Principles of Physics and General College Physics problem 8.43: Energy to bring centrifuge motor with moment of inertia 3.75 * 10^-2 kg m^2 to 8250 rpm from rest.
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11:01:23 The KE of a rotating object is KE = .5 I omega^2, where I is the moment of inertia and omega the angular velocity. Since I is given in standard units of kg m^2, the angular velocity should be expressed in the standard units rad / sec. Since 8250 rpm = (8250 rpm) * (pi / 30 rad/sec) / rpm = 860 rad/sec, approx.. The initial KE is 0, and from the given information the final KE is KE_f = .5 I omega_f ^ 2 = .5 * 3.75 * 10^-2 kg m^2 * (860 rad/sec)^2 = 250 pi^2 kg m^2 / sec^2 = 14000 Joules.
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11:01:24 Query gen problem 8.58 Estimate KE of Earth around Sun (6*10^24 kg, 6400 km rad, 1.5 * 10^8 km orb rad) and about its axis. What is the angular kinetic energy of the Erath due to its rotation about the Sun? What is the angular kinetic energy of the Earth due to its rotation about its axis?
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11:01:25 ** The circumference of the orbit is 2pi*r = 9.42*10^8 km. We divide the circumference by the time required to move through that distance to get the speed of Earth in its orbit about the Sun: 9.42 * 10^8 km / (365days * 24 hrs / day * 3600 s / hr) =29.87 km/s or 29870 m/s. Dividing the speed by the radius we obtain the angular velocity: omega = (29.87 km/s)/ (1.5*10^8 km) = 1.99*10^-7 rad/s. From this we get the angular KE: KE = 1/2 mv^2 = 1/2 * 6*10^24 kg * (29870 m/s)^2 = 2.676*10^33 J. Alternatively, and more elegantly, we can directly find the angular velocity, dividing the 2 pi radian angular displacement of a complete orbit by the time required for the orbit. We get omega = 2 pi rad / (365days * 24 hrs / day * 3600 s / hr) = 1.99 * 10^-7 rad/s. The moment of inertia of Earth in its orbit is M R^2 = 6 * 10^24 kg * (1.5 * 10^11 m)^2 = 1.35 * 10^47 kg m^2. The angular KE of the orbit is therefore KE = .5 * I * omega^2 = .5 * (1.35 * 10^47 kg m^2) * (1.99 * 10^-7 rad/s)^2 = 2.7 * 10^33 J. The two solutions agree, up to roundoff errors. The angular KE of earth about its axis is found from its angular velocity about its axis and its moment of inertia about its axis. The moment of inertia is I=2/5 M r^2=6*10^24kg * ( 6.4 * 10^6 m)^2 = 9.83*10^37kg m^2. The angular velocity of the Earth about its axis is 1 revolution / 24 hr = 2 pi rad / (24 hr * 3600 s / hr) = 7.2 * 10^-5 rad/s, very approximately. So the angular KE of Earth about its axis is about KE = .5 I omega^2 = .5 * 9.8 * 10^37 kg m^2 * (7.2 * 10^-5 rad/s)^2 = 2.5 * 10^29 Joules. **
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11:01:27 Query problem 8.60 uniform disk at 2.4 rev/sec; nonrotating rod of equal mass, length equal diameter, dropped concentric with disk. Resulting angular velocity?
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11:01:28 ** The moment of inertia of the disk is I = 2/5 M R^2; the moment of inertia of the rod about its center is 1/12 M L^2. The axis of rotation of each is the center of the disk so L = R. The masses are equal, so we find that the moments of inertia can be expressed as 2/5 M R^2 and 1/12 M R^2. The combined moment of inertia is therefore 2/5 M R^2 + 1/12 M R^2 = 29/60 M R^2, and the ratio of the combined moment of inertia to the moment of the disk is ratio = (29/60 M R^2) / (2/5 M R^2) = 29/60 / (2/5) = 29/60 * 5/2 = 145 / 120 = 29 / 24. Since angular momentum I * omega is conserved an increase in moment of inertia I results in a proportional decrease in angular velocity omega so we end up with final angular velocity = 24 / 29 * initial angular velocity = 24 / 29 * 2.4 rev / sec = 2 rev/sec, approximately.
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11:01:29 Univ. 10.64 (10.56 10th edition). disks 2.5 cm and .8 kg, 5.0 cm and 1.6 kg, welded, common central axis. String around smaller, 1.5 kg block suspended. Accel of block? Then same bu wrapped around larger.
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11:01:30 ** The moment of inertia of each disk is .5 M R^2; the block lies at perpendicular distance from the axis which is equal to the radius of the disk to which it is attached. So the moment of inertia of the system, with block suspended from the smaller disk, is I = .5 (.8 kg) * ( .025 m)^2 + .5 * 1.6 kg * (.05 m)^2 + (1.5 kg * .025 m)^2= .0032 kg m^2 approx. The 1.5 kg block suspended from the first disk results in torque tau = F * x = .025 m * 1.5 kg * 9.8 m/s^2 = .37 m N approx. The resulting angular acceleration is alpha = tau / I = .37 m N / (.0032 kg m^2) = 115 rad/s^2 approx. The acceleration of the block is the same as the acceleration of a point on the rim of the wheel, which is a = alpha * r = 115 rad/s^2 * .025 m = 2.9 m/s^2 approx. The moment of inertia of the system, with block suspended from the larger disk, is I = .5 (.8 kg) * ( .025 m)^2 + .5 * 1.6 kg * (.05 m)^2 + (1.5 kg * .05 m)^2= .006 kg m^2 approx. The 1.5 kg block suspended from the first disk results in torque tau = F * x = .05 m * 1.5 kg * 9.8 m/s^2 = .74 m N approx. The resulting angular acceleration is alpha = tau / I = .74 m N / (.006 kg m^2) = 120 rad/s^2 approx. The acceleration of the block is the same as the acceleration of a point on the rim of the wheel, which is a = alpha * r = 120 rad/s^2 * .05 m = 6 m/s^2 approx. **
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YɮzDTNglvһ assignment #033 骶͌Άrzvc֡ Physics I 08-06-2008
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21:49:08 Query modeling simple harmonic motion with a reference circle. In what sense can we say that the motion of a pendulum is modeled by the motion of a point moving at constant velocity around a reference circle? Be specific.
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RESPONSE --> If it were moving in a circle the distance traveled would be the circumference of the circle, which is 2pi r
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21:49:13 GOOD STUDENT ANSWER: A point moving around a circle can be represented by two perpendicular lines whose intersection is that point point of constant velocity. The vertical line then is one that moves back and forth, which can be sychronized to the oscillation of the pendulum.
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21:49:44 At what point(s) in the motion a pendulum is(are) its velocity 0?
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RESPONSE --> A pendulum is at velocity 0m/sec when it is about to change direction, which is at both extremes.
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21:50:31 GOOD STUDENT ANSWER: The pendulum has two points of v= 0. One at each end as it briefly comes to a stop to benin swinging in the opposite direction. At what point(s) in the motion a pendulum is(are) its speed a maximum?
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RESPONSE --> The pendulum is at maximum speed when it travels half the distance to the other end. This is because acceleratiom due to gravity is working for the pendulum
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21:50:51 GOOD STUDENT ANSWER: The mid point velocity of the pendulum represents its greatest speed since it begins at a point of zero and accelerates by gravity downward to equilbrium, where it then works against gravity to finish the oscillation. GOOD STUDENT DESRIPTION OF THE FEELING: At the top of flight, the pendulum 'stops' then starts back the other way. I remeber that I used to love swinging at the park, and those large, long swings gave me such a wonderful feeling at those points where I seemed tostop mid-air and pause a fraction of a moment.Then there was that glorious fall back to earth. Too bad it makes me sick now. That was how I used to forget all my troubles--go for a swing. *&*& INSTRUCTOR COMMENT: That extreme point is the point of maximum acceleration. **
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21:51:27 How does the maximum speed of the pendulum compare with the speed of the point on the reference circle?
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RESPONSE --> If a circle is moving in a vertical motion, the maximum speed is at the bottom of the circle when again acceleration is working for the system
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21:51:46 ** At the equilibrium points the pendulum is moving in the same direction and with the same speed as the point on the reference circle. University Physics Note: You can find the average speed by integrating the speed function, which is the absolute value of the velocity function, over a period and then dividing by the period (recall from calculus that the average value of a function over an interval is the integral divided by the length of the interval). You find rms speed by finding the average value of the squared velocity and taking the square root (this is the meaning of rms, or root-mean-square). **
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21:52:04 How can we determine the centripetal acceleration of the point on the reference circle?
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RESPONSE --> centripetal acceleration is defined to be aCemt = v^2 / r
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21:52:07 ** Centripetal acceleration is v^2 / r. Find the velocity of a point on the reference circle (velocity = angular velocity * radius). **
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21:52:10 Query gen phy problem 9.12 30 kg light supported by wires at 37 deg and 53 deg with horiz. What is the tension in the wire at 37 degrees, and what is the tension in the other wire?
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21:52:11 ** The given solution is for a 30 kg light. You should be able to adapt the details of this solution to the 33 kg traffic light in the current edition: The net force on the light is 0. This means that the net force in the vertical direction will be 0 and likewise for the net force in the horizontal direction. We'll let the x axis be horizontal and the y axis vertical and upward. Let T1 be the tension in the 37 deg wire and T2 the tension in the 53 deg wire. Assuming that the 37 deg is with the negative x axis then T1 acts at the angle 180 deg - 37 deg = 143 deg. Gravity exerts a downward force of 30kg * 9.8 m/s^2 = 294N. The x and y components of the forces are as follows: x y weight 0 -294 N T1 T1 cos(143 deg) T1 sin(143 deg) T2 T2 cos(53 deg) T2 sin(53 deg) The net force in the x direction is T1 cos(143 deg) + T2 cos(53 deg) = -.8 T1 + .6 T2 The net force in the y direction is T1 sin(143 deg) + T2 sin(53 deg) - 294 N = .6 T1 + .8 T2 - 294 N. These net forces are all zero so -.8 T1 + .6 T2 = 0 and .6 T1 + .8 T2 - 294 N = 0. Solving the first equation for T1 in terms of T2 we obtain T1 = .75 T2. Plugging this result into the first equation we get .6 ( .75 T2) + .8 T2 - 294 N = 0 which we rearrange to get 1.25 T2 = 294 N so that T2 = 294 N / 1.25 = 235 N approx. Thus T1 = .75 T2 = .75 * 235 N = 176 N approx.. **
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21:52:12 Query problem 9.19 172 cm person supported by scales reading 31.6 kg (under feet) and 35.1 kg (under top of head).
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21:52:14 ****The solution given here is for a person 170 cm tall, rather than 172 cm tall. You should be able to adapt the given solution to the 172 cm height; all distances will increase by factor 172 / 170 = 86 / 85, a little more than 1%: The center of gravity is the position for which the net torque of the person is zero. If x represents the distance of this position from the person's head then this position is also 170 cm - x from the person's feet. The 35.1 kg reading indicates a force of 35.1 kg * 9.8 m/s^2 = 344 N and the 31.6 kg reading indicates a force of 31.6 kg * 9.8 m/s^2 = 310 N, both results approximate. About the point x cm from the head we then have the following, assuming head to the left and feet to the right: }torque of force supporting head = -344 N * x torque of force supporting feet = 310 N * (170 cm - x). Net torque is zero so we have -344 N * x + 310 N * (170 cm - x) = 0. We solve for x: -344 N * x + 310 N * 170 cm - 310 N * x = 0 -654 N * x = -310 N * 170 cm x = 310 N * 170 cm / (654 N) = 80.5 cm. The center of mass is therefore 80.5 cm from the head, 89.5 cm from the feet. **
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21:52:15 Principles of Physics and General College Physics Problem 9.2: 58 kg on diving board 3.0 m from point B and 4.0 m from point A; torque about point B:
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21:52:16 The torque exerted by the weight of the 58 kg person is torque = moment arm * force = 3.0 meters * (58 kg * 9.8 m/s^2) = 3.0 meters * 570 N = 1710 meter * newtons.
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21:52:18 Principles of Physics and General College Physics Problem 9.30: weight in hand 35 cm from elbow joint, 2.0 kg at CG 15 cm from joint, insertion 6.0 cm from joint. What weight can be held with 450 N muscle force?
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21:52:20 Query gen problem 9.32 arm mass 3.3 kg, ctr of mass at elbow 24 cm from shoulder, deltoid force Fm at 15 deg 12 cm from shoulder, 15 kg in hand. Give your solution:
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21:52:21 **The total torque about the shoulder joint is zero, since the shoulder is in equilibrium. Also the net vertical force on the arm is zero, as is the net horizontal force on the arm. The 3.3 kg mass of the arm experiences a downward force from gravity of w = 3.3 kg * 9.8 m/s^2 = 32 N, approx. At 24 cm from the joint the associated torque is 32 N * .24 m = 8 m N, approx. THe 15 kg in the hand, which is 60 cm from the shoulder, results in a torque of 15 kg * 9.8 m/s^2 * .60 m = 90 m N, approx. }The only other force comes from the deltoid, which exerts its force at 15 degrees from horizontal at a point 12 cm from the joint. If F is the force exerted by the deltoid then the resulting torque is F * sin(15 deg) * .12 m = .03 F, approx.. If we take the torques resulting from gravitational forces as negative and the opposing torque of the deltoid as positive then we have - 8 m N - 90 m N + .03 F = 0 (sum of torques is zero), which we easily solve to obtain F = 3300 N. This 3300 N force has vertical and horizontal components 3300 N * sin(15 deg) = 800 N approx., and 3300 N * cos(15 deg) = 3200 N approx.. The net vertical force on the arm must be zero. There is a force of 800 N (vert. comp. of deltoid force) pulling up on the arm and 32 N (gravitational force) pulling down, which would result in a net upward vertical force of 768 Newtons, so there must be another force of 768 N pulling downward. This force is supplied by the reaction force in the shoulder as the head of the humerus is restrained by the 'socket' of the scapula and the capsule of ligaments surrounding it. The net horizontal force must also be zero. The head of the humerus is jammed into the scapula by the 3200 N horizontal force, and in the absence of such things as osteoporosis the scapula and capsule easily enough counter this with an equal and opposite force. **
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21:52:24 Univ. 11.62 (11.56 10th edition). .036 kg ball beneath .024 kg ball; strings at angles 53.1 deg and 36.9 deg to horiz rod suspended by strings at ends, angled strings .6 m apart when joining rod, .2 m from respective ends of rod. Tension in strings A, B, C, D, E, F (lower ball, upper ball, 53 deg, 37 deg, 37 deg end of rod, 53 deg end).
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21:52:26 ** Cord A supports the .0360 kg ball against the force of gravity. We have T - m g = 0 so T = m g = .0360 kg * 9.8 m/s^2 = .355 N. The second ball experiences the downward .355 N tension in string A, the downward force .0240 kg * 9.8 m/s^2 = .235 N exerted by gravity and the upward force Tb of tension in string B so since the system is in equilibrium Tb - .355 n - .235 N = - and Tb = .59 N. If Tc and Td are the tensions in strings C and D, since the point where strings B, C and D join are in equilibrium we have Tcx + Tdx + Tbx = 0 and Tcy + Tdy + Tby = 0. Noting that strings C and D respectively make angles of 53.1 deg and 143.1 deg with the positive x axis we have Tby = =.59 N and Tbx = 0. Tcx = Tc cos(53.1 deg) = .6 Tc Tcy = Tc sin(53.1 deg) = .8 Tc Tdx = Td cos(143.1 deg) = -.8 Td Tdy = Td sin(143.1 deg) = .6 Td. So our equations of equilibrium become .6 Tc - .8 Td = 0 .8 Tc + .6 Td - .59 N = 0. The first equation tells us that Tc = 8/6 Td = 4/3 Td. Substituting this into the second equation we have .8 (4/3 Td) + .6 Td - .59 N = 0 1.067 Td + .6 Td = .59 N 1.667 Td = .59 N Td = .36 N approx. so that Tc = 4/3 Td = 4/3 (.36 N) = .48 N approx.. Now consider the torques about the left end of the rod. We have torques of -(.200 m * Td sin(36.9 deg)) = -.200 m * .36 N * .6 = -.043 m N (note that this torque is clockwise, therefore negative). -(.800 m * Tc sin(53.1 deg) = -.800 m * .48 N * .8 = -.31 m N and 1.0 m * Tf, where Tf is the tension in string F. Total torque is 0 so -.043 m N - .31 m N + 1.0 m * Tf = 0 and Tf = .35 N approx.. The net force on the entire system is zero so we have Te + Tf - .59 N = 0 or Te = .59 N - Tf = .59 N - .35 N = .24 N. **
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