course PHY 231 yĴ~٤sczw˵assignment #021 骶͌Άrzvc֡ Physics I 08-04-2008
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02:08:30 Explain how to obtain the final speed and direction of motion of a projectile which starts with known velocity in the horizontal direction and falls a known vertical distance, using the analysis of vertical and horizontal motion and vectors.
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RESPONSE --> The final speed is obtained using the following equation Projectile v = square root (Vx^2 + Vy^2) To find the angle use tan y/x. In the horizontal direction the velocity is constant throughout which means that the acceleration is 0m/sec^2. In the vertical direction the acceleration due to gravity is 9.8m/sec^2.
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02:09:13 ** The horizontal velocity is unchanging so the horizontal component is always equal to the known initial horizontal velocity. The vertical velocity starts at 0, with acceleration thru a known distance at 9.8 m/s^2 downward. The final vertical velocity is easily found using the fourth equation of motion. We therefore know the x (horizontal) and y (vertical) components of the velocity. Using the Pythagorean Theorem and arctan (vy / vx) we find the speed and direction of the motion. **
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RESPONSE --> I clicked enter response on accident so I do not know if my answer went through. However, this is very similar to what I said,
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02:09:50 Give at least three examples of vector quantities for which we might wish to find the components from magnitude and direction. Explain the meaning of the magnitude and the direction of each, and explain the meaning of the vector components.
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RESPONSE --> Forces, velocities,and displacment. These are all vectors and are treated as such.
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02:10:02 ** GOOD STUDENT RESPONSE: Examples might include: A force acting on an object causing it to move in an angular direction. A ball falling to the ground with a certain velocity and angle. A two car collision; velocity and momentum are both vector quantities and both important for analyzing the collision.. The magnitude and directiohn of the relsultant is the velocity and direction of travel. The vector components are the horizontal and vertical components that would produce the same effect as the resultant.
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RESPONSE --> ok
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o^MЄì_ assignment #022 骶͌Άrzvc֡ Physics I 08-04-2008
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03:14:11 Query gen phy 7.19 95 kg fullback 4 m/s east stopped in .75 s by tackler due west
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03:14:14 ** We'll take East to be the positive direction. The origional magnitude and direction of the momentum of the fullback is p = m * v1 = 115kg (4m/s) = 380 kg m/s. Since velocity is in the positive x direction the momentum is in the positive x direction, i.e., East. The magnitude and direction of the impulse exerted on the fullback will therefore be impulse = change in momentum or impulse = pFinal - pInitial = 0 kg m/s - 380 kg m/s = -380 kg m/s. Impulse is negative so the direction is in the negative x direction, i.e., West. Impulse = Fave * `dt so Fave = impulse / `dt. Thus the average force exerted on the fullback is Fave = 'dp / 'dt = -380 kg m/s /(.75s) = -506 N The direction is in the negative x direction, i.e., West. The force exerted on the tackler is equal and opposite to the force exerted on the fullback. The force on the tackler is therefore + 506 N. The positive force is consistent with the fact that the tackler's momentum change in positive (starts with negative, i.e., Westward, momentum and ends up with momentum 0). The iimpulse on the tackler is to the East. **
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C|Fw assignment #023 骶͌Άrzvc֡ Physics I 08-04-2008
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03:32:47 Query gen phy 7.27 bumper cars 450 kg at 4.5 m/s, 550 kg at 3.7 m/s, collision from back, elastic
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03:32:49 ** For an elastic collision we have m1 v1 + m2 v2 = m1 v1' + m2 v2' and v2 - v1 = -( v2' - v1'). We substitute m1, v1, m2 and v2 to obtain 450 kg * 4.5 m/s + 550 kg * 3.7 m/s = 450 kg * v1 ' + 550 kg * v2 ', or 4060 kg m/s = 450 kg * v1 ' + 550 kg * v2 ' . Dividing by 10 and by kg we have 406 m/s = 45 v1 ' + 55 v2 '. We also obtain 3.7 m/s - 4.5 m/s = -(v2 ' - v1 ' ) or v1 ' = v2 ' - .8 m/s. Substituting this into 406 m/s = 45 v1 ' + 55 v2 ' we obtain 406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' . We easily solve for v2 ' obtaining v2 ' = 4.42 m/s. This gives us v1 ' = 4.42 m/s - .8 m/s = 3.62 m/s. Checking to be sure that momentum is conserved we see that the after-collision momentum is pAfter = 450 kg * 3.62 m/s + 550 kg * 4.42 m/s = 4060 m/s. The momentum change of the first car is m1 v1 ' - m1 v1 = 450 kg * 3.62 m/s - 450 kg * 4.5 m/s = - 396 kg m/s. The momentum change of the second car is m2 v2 ' - m2 v2 = 550 kg * 4.42 m/s - 550 kg * 3.7 m/s = + 396 kg m/s. Momentum changes are equal and opposite. NOTE ON SOLVING 406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' FOR v2 ': Starting with 406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' use the Distributive Law to get 406 m/s = 45 v2 ' - 36 m/s + 55 v2 ' then collect the v2 ' terms to get 406 m/s = -36 m/s + 100 v2 '. Add 36 m/s to both sides: 442 m/s = 100 v2 ' so that v2 ' = 442 m/s / 100 = 4.42 m/s. *
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03:33:28 Univ. 3.48. (not in 11th edition) ball 60 deg wall 18 m away strikes 8 m higher than thrown. What are the Init speed of the ball and the magnitude and angle of the velocity at impact?
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RESPONSE --> I have the 11th edition so I am not able to solve this problem.
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03:33:31 ** We know the following: For y motion `dsy = + 8 m, ay = -g = - 9.8 m/s^2 and v0y = v0 sin(60 deg) = .5 v0. For x motion `dsx = 18 m, ax = 0 and v0x = v0 cos(60 deg) = .87 v0, approx. Assuming a coordinate system where motion starts at the origin: The equation of motion in the x direction is thus x = .5 v0 * t and the equation of y motion is y = .87 v0 t - .5 g t^2. We know x and y at impact and we know g so we could solve these two equations simultaneously for v0 and t. We begin by eliminating t from the two equations: x = .5 v0 * t so t = 2 x / v0. Substituting this expression for t in the second equation we obtain y = .87 v0 * (2 x / v0) - .5 g ( 2 x / v0) ^ 2. Multiplying both sides by v0^2 we obtain v0^2 y = .87 v0^2 ( 2 x) - .5 g * 4 x^2. Bringing all the v0 terms to the left-hand side we have v0^2 y - 1.73 v0^2 x = -2 g * x^2. Factoring v0 we have v0^2 ( y - 1.73 x) = -2 g x^2 so that v0 = +-sqrt(-2 g x^2 / ( y - 1.73 x) ) ) . Since we know that y = 8 m when x = 18 m we obtain = +- sqrt( -9.8 m/s^2 * (18 m)^2 / ( 8 m - 1.73 * 18 m) ) = +-sqrt(277 m^2 / s^2) = +-16.7 m/s, approx.. We choose the positive value of v0, since the negative value would have the projectile moving 'backward' from its starting point. Substituting this value into t = 2 x / v0 and recalling that our solution applies to the instant of impact when x = 16 m and y = 8 m we obtain t = 2 * 18 m / (16.7 m/s) = 2.16 s. Alternatively we can solve the system for v0 less symbolically and perhaps gain different insight into the meaning of the solution. Starting with the equations x = .5 v0 * t and y = .87 v0 t - .5 g t^2 we see that impact occurs when x = .5 v0 t = 18 m so that t = 18 m / (.5 v0) = 36 m / v0. At this instant of impact y = 8 m so substituting this and the t just obtained into the equation of motion for y we get y = .87 v0 (36 m / v0 ) - .5 g (36 m / v0 )^2 = 8 m. The equation .87 v0 (36 m / v0 ) - .5 g (36 m / v0 )^2 = 8 m is easily solved for v0, obtaining v0 = 16.7 m/s. With this initial velocity we again confirm that t = 2.16 sec at impact. Note that at t = 2.16 sec we get y = 14.4 m/s * 2.16 s - 4.9 m/s^2 * (2.16 s)^2 = 8 m, within roundoff error, confirming this solution. We need the magnitude and direction of the velocity at impact. We therefore need the x and y components of the velocity at the t = 2.16 sec instant. At this instant we have x and y velocities vx = dx/dt = .5 v0 = 8.35 m/s and vy = dy/dt = 14.4 m/s - 9.8 m/s^2 * t = 14.4 m/s - 9.8 m/s^2 * 2.16 s = -5.6 m/s, approx. The velocity at impact therefore has magnitude sqrt( (8.25 m/s)^2 + (-5.6 m/s)^2 ) = 10 m/s, approx. and the angle is arctan(vy/vx) = arctan(-5.6 / 8.25) = -34 deg, approx. At impact the object is moving at 10 m/s and at 34 deg below horizontal (i.e., it's on its way back down). **
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R靇kץ̥ƽ ЬH assignment #024 骶͌Άrzvc֡ Physics I 08-04-2008
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04:38:11 Why was it necessary to let the string go slack at the top of the circle in order to get the desired results?
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RESPONSE --> If it was straight it would by far pass the revolution and continue at its velocity
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04:38:30 ** If the string goes slack just at the instant the weight reaches the 'top' of its circular path then we are assured that the centripetal acceleration is equal to the acceleration of gravity. If there is tension in the string then the weight is being pulled downward and therefore toward the center by a force that exceeds its weight. If the string goes slack before the weight reaches the top of its arc then the path isn't circular and our results won't apply to an object moving in a circular arc. **
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04:39:08 Why do you expect that, if the string is released exactly at the top of the circle, the initial velocity of the washer will be horizontal?
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RESPONSE --> It will not have any horizontal direction so it will only have a vertical velocity
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04:39:17 ** The direction of an object moving in a circular arc is perpendicular to a radial line (i.e., a line from the center to a point on the circle). When the object is at the 'top' of its arc it is directly above the center so the radial line is vertical. Its velocity, being perpendicular to this vertical, must be entirely in the horizontal direction. **
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04:39:42 What is the centripetal acceleration of the washer at the instant of release, assuming that it is released at the top of its arc and that it goes slack exactly at this point, and what was the source of this force?
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RESPONSE --> 9.8m/sec^2 which is the same as the acceleration due to gravity.
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04:39:46 ** Under these conditions, with the string slack and not exerting any force on the object, the centripetal acceleration will be equal to the acceleration of gravity. **
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04:42:51 Query principles of physics and general college physics 7.02. Delivery truck 18 blocks north, 10 blocks east, 16 blocks south. What is the final displacement from the origin?
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RESPONSE --> Move this into x and y components. X is east Y is North and South X is 10 blocks east Y is 2 blocks north Find the final displacement using vector sums square root (x^2 y^2) sq. root ( 10^2 + 2^2) Sq root (104) 10.198 blocks 11.13 degrees North of east I still need to find direction tan y /x tan 2 / 10
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04:43:29 The final position of the truck is 2 blocks south and 10 blocks east. This is equivalent to a displacement of +10 blocks in the x direction and -2 blocks in the y direction. The distance is therefore sqrt( (10 blocks)^2 + (-2 blocks)^2 ) = sqrt( 100 blocks^2 + 4 blocks^2) = sqrt(104 blocks^2) = sqrt(104) * sqrt(blocks^2) = 10.2 blocks. The direction makes and angle of theta = arcTan(-2 blocks / (10 blocks) ) = arcTan(-.2) = -12 degrees with the positive x axis, as measured counterclockwise from that axis. This puts the displacement at an angle of 12 degrees in the clockwise direction from that axis, or 12 degrees south of east.
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RESPONSE --> I used a different direction
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04:43:32 Query principles of physics and general college physics 7.18: Diver leaves cliff traveling in the horizontal direction at 1.8 m/s, hits the water 3.0 sec later. How high is the cliff and how far from the base does he land?
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04:43:34 The diver's initial vertical velocity is zero, since the initial velocity is horizontal. Vertical velocity is characterized by the acceleration of gravity at 9.8 m/s^2 in the downward direction. We will choose downward as the positive direction, so the vertical motion has v0 = 0, constant acceleration 9.8 m/s^2 and time interval `dt = 3.0 seconds. The third equation of uniformly accelerated motion tells us that the vertical displacement is therefore vertical motion: `ds = v0 `dt + .5 a `dt^2 = 0 * 3.0 sec + .5 * 9.8 m/s^2 * (3.0 sec)^2 = 0 + 44 m = 44 m. The cliff is therefore 44 m high. The horizontal motion is characterized 0 net force in this direction, resulting in horizontal acceleration zero. This results in uniform horizontal velocity so in the horizontal direction v0 = vf = vAve. Since v0 = 1.8 m/s, vAve = 1.8 m/s and we have horizontal motion: `ds = vAve * `dt = 1.8 m/s * 3.0 s = 5.4 meters.
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04:43:36 Gen phy 3.13 A 44 N at 28 deg, B 26.5 N at 56 deg, C 31.0 N at 270 deg. Give your solution to the problem.
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04:43:37 ** The solution given here is for a previous edition, in which the forces are Force A of 66 at 28 deg Force B of B 40 at 56 deg Force C of 46.8 at 270 deg These forces are very close to 2/3 as great as the forces given in the current edition, and all correct results will therefore be very close to 2/3 as great as those given here. Calculations to the nearest whole number: A has x and y components Ax = 66 cos(28 deg) = 58 and Ay = 66 sin(28 deg) = 31 Bhas x and y components Bx = 40 cos(124 deg) = -22 and By = 40 sin(124 deg) = 33 C has x and y components Cx = 46.8 cos(270 deg) = 0 and Cy = 46.8 sin(270 deg) = -47 A - B + C therefore has components Rx = Ax-Bx+Cx = 58 - (-22) + 0 = 80 and Ry = Ay - By + Cy = 31-33-47=-49, which places it is the fourth quadrant and gives it magnitude `sqrt(Rx^2 + Ry^2) = `sqrt(80^2 + (-49)^2) = 94 at angle tan^-1(Ry / Rx) = tan^-1(-49/53) = -32 deg or 360 deg - 32 deg = 328 deg. Thus A - B + C has magnitude 93 at angle 328 deg. B-2A has components Rx = Bx - 2 Ax = -22 - 2 ( 58 ) = -139 and Ry = By - 2 Ay = 33 - 2(31) = -29, placing the resultant in the third quadrant and giving it magnitude `sqrt( (-139)^2 + (-29)^2 ) = 142 at angle tan^-1(Ry / Rx) or tan^-1(Ry / Rx) + 180 deg. Since x < 0 this gives us angle tan^-1(-29 / -139) + 180 deg = 11 deg + 180 deg = 191 deg. Thus B - 2 A has magnitude 142 at angle 191 deg. Note that the 180 deg is added because the angle is in the third quadrant and the inverse tangent gives angles only in the first or fourth quandrant ( when the x coordinate is negative we'll be in the second or third quadrant and must add 180 deg). **
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04:43:51 Univ. 3.58. (This problem has apparently been eliminated from recent editions, due to the now policitally incorrect nature of the device being thrown. The problem is a very good one and has been edited to eliminate politically incorrect references). Good guys in a car at 90 km/hr are following
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04:44:20 bad guys driving their car, which at a certain instant is 15.8 m in front of them and moving at a constant 110 km/hr; an electronic jamming device is thrown by the good guys at 45 deg above horizontal, as they observe it. This device must land in the bad guy's car. With what speed must the device be thrown relative to the good guys, and with what speed relative to the ground?
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04:44:34 ** The device is thrown at velocity v0 at 45 deg, giving it v0y = .71 v0 and v0x = .71 v0. The device will return to its original vertical position so we have `dsy = 0. Using `dsy = v0y `dt + .5 g `dt^2 with `dsy = 0 and assuming the upward direction to be positive we obtain v0y `dt + .5 (-g) `dt^2 = 0 so that `dt = 0 or `dt = - 2 * v0y / (-g) = 2 * 71 v0 / g. In time `dt the horizontal displacement relative to the car will be `dsx = v0x `dt + ax `dt; since acceleration ax in the x direction and v0x = .71 v0 is zero we have `dsx = .71 v0 * `dt. We also know that relative to the first car the second is moving at 20 km / hr = 20,000 m / (3600 sec) = 5.55 m/s, approx.; since its initial position is 15.8 m in front of the first car we have `dsx = 15.8 m + 5.55 m/s * `dt. To keep the equations symbolic we use x0Relative and vRelative for the relative initial position and velocity of the second car with respect to the first. We thus have three equations: `dt = 2 * .71 v0 / g = 1.42 v0 / g. `dsx = .71 v0 * `dt `dsx = x0Relative + v0Relative * `dt. This gives us three equations in the variables v0, `dt and `dsx, which we reduce to two by substituting the expression -2 to obtain: `dsx = .71 v0 * 1.42 v0 / g = v0^2 / g `dsx = x0Relative + v0Relative * 1.42 v0 / g. Setting the right-hand sides equal we have v0^2 / g = x0Relative + v0Relative * 1.42 v0 / g, or v0^2 - v0Relative * 1.42 v0 - g * x0Relative = 0. We get v0 = [1.42 v0Relative +-sqrt( (1.42 v0Relative)^2 - 4 * (-g * x0Relative) ) ] / 2 = [1.42 * v0Relative +-sqrt( (1.42 * v0Relative)^2 + 4 * g * x0Relative) ] / 2. Substituting 5.55 m/s for v0Relative and 15.8 m for x0Relative we get [1.42 * 5.55 m/s +-sqrt( (1.42 *5.55 m/s)^2 + 4 * 9.8 m/s^2 *15.8 m) ] / 2 = 17 m/s or -9.1 m/s, approx.. We conclude that the initial velocity with respect to the first case must be 17 m/s. Checking this we see that the device will have initial x and y velocities 7.1 * 17 m/s = 12 m/s, approx., and will therefore stay aloft for 2 * 12 m/s / (9.8 m/s^2) = 2.4 sec, approx.. It will therefore travel 2.4 sec * 12 m/s = 28 m, approx. in the horizontal direction relative to the first car. During this time the second car will travel about 5.55 m/s * 2.4 sec = 13 m, approx., resulting in relative position 15.8 m + 13 m = 28.8 m with respect to the first. This is reasonably close to the 28 m obtained from the motion of the projectile. Correcting for roundoff errors will result in precise agreement. **
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SԀwfe٢ assignment #025 骶͌Άrzvc֡ Physics I 08-04-2008
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04:57:01 principles of physics and gen phy 4.26 free-body diagram of baseball at moment hit, flying toward outfield gen phy list the forces on the ball while in contact with the bat, and describe the directions of these forces
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04:57:03 ** Gravity exerts a downward force equal to the weight of the ball. While in contact with the ball, and only while i contact, the bat exerts a normal force, which pushes outward along a line originating from the central axis of the bat. This force is perpendicular to the surface of the bat at the point of contact. Unless the direction of the ball is directly toward the center of the bat, which will not be the case if the ball is hit at an upward angle by a nearly level swing, there will also be a frictional force between bat and ball. This frictional force will be parallel to the surface of the bat and will act on the ball in the 'forward' direction. COMMON STUDENT ERROR: The gravitational force and the force exerted by the ball on the bat are equal and opposite. The force of the bat on the ball and the gravitational force are not equal and opposite, since this is not an equilibrium situation--the ball is definitely being accelerated by the net force, so the net force is not zero. **
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04:57:04 gen phy list the forces on the ball while flying toward the outfield, and describe the directions of these forces
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04:57:06 **After impact the forces are gravity, which is constant and in the y direction, and air resistance. The direction of the force of air resistance is opposite to the direction of motion. The direction of motion is of course constantly changing, and the magnitude of the force of air resistance depends on the speed of the ball with respect to the air, which is also changing. **
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04:57:08 gen phy give the source of each force you have described
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04:57:09 ** The gravitational force is the result of the gravitational attraction between the ball and the Earth. The normal force is the result of the elastic compression of bat and ball. The frictional force is due to a variety of phenomena related to the tendency of the surfaces to interlock (electromagnetic forces are involved) and to encounter small 'bumps' in the surfaces. **
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04:57:10 gen phy what is the direction of the net force on the ball while in contact with the bat?
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04:57:11 ** The normal force will vary from 0 at the instant contact begins to a maximum at the instant of greatest compression, and back to 0 at the instant contact ceases. So there is no single normal force. However we can represent 'the' normal force as the average normal force. The gravitational force will remain constant; the frictional force will vary along with the normal force, and we will speak here of the average frictional force.The average normal force will be the greatest force, much greater than friction or gravity. The frictional force will likely also exceed the gravitational force. The y component of the normal force will overwhelm the y components of the frictional force and the gravitational force, both of which are downward, giving us a net y component slightly less than the y component of the normal force. The x component of the normal force will be reinforced by the x component of the frictional force, making the x component of the net force a bit greater than the x component of the normal force. This will result in a net force that is 'tilted' forward and slightly down from the normal force. Note that the frictional force will tend to 'spin' the baseball but won't contribute much to the translational acceleration of the ball. This part is a topic for another chapter. **
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04:57:13 gen phy what is the net force on the ball while flying toward the outfield?
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04:57:14 ** The net force will consist of the downward gravitational force and the force of air resistance opposing the motion. If the ball is rising the y component of the air resistance will be in the downward direction, reinforcing the gravitational force and giving a net downward y component slightly exceeding that of gravity. If the ball is falling the y component will be in the upward direction, opposing the gravitational force and giving a net downward y component slightly less than that of gravity. In either case the x component will be in the direction opposite to the motion of the ball, so the net force will be directed mostly downward but also a bit 'backward'. There are also air pressure forces related to the spinning of the ball; the net force exerted by air pressure causes the path of the ball to curve a bit, but these forces won't be considered here. **
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05:04:27 Univ. 5.88 (5.84 10th edition). Elevator accel upward 1.90 m/s^2; 28 kg box; coeff kin frict 0.32. How much force to push at const speed?
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05:05:14 STUDENT SOLUTION AND INSTRUCTOR COMMENT: The magnitude of kinetic friction force is fk = mu-sub k * N. First we add the 1.9 to 9.8 and get 11.7 as the acceleration and times that by the 28 kg and get 327.6 as the force so plugging in we get fk = 0.32 * 327.6 = 104.8 N. ** Good. The net force Fnet on the box is Fnet = m a = 1.90 m/s^2 * 28 kg. The net force is equal to the sum of the forces acting on the box, which include the weight mg acting downward and the force of the floor on the box acting upward. So we have Fnet = Ffloor - m g = m a. Thus Ffloor = m g + m a = 28 kg * 9.8 m/s^2 + 28 kg * 1.90 m/s^2 = 28 kg * 11.7 m/s^2 = 330 N, approx. Being pushed at constant speed the frictional force is f = `mu * N, where N is the normal force between the box and the floor. So we have f = .32 * 330 Newtons = 100 N, approx. **
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RESPONSE --> How can you add the acceleration of 1.9 to 9.8 when 1.9 is moving upward?
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aw~ޡΏ{q룆 assignment #026 骶͌Άrzvc֡ Physics I 08-04-2008
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05:16:36 gen phy and principles of phy 4.36: If the coefficient of kinetic friction is .30, how much force is required to push a 35 lb crate across the floor at constant speed?
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05:16:37 If the crate is moving then the force exerted by friction is .30 times the normal force between it and the floor. If the push is horizontal, then the only horizontal forces acting on the crate are the downward force of gravity and the upward force exerted by the floor. Since the crate is not accelerating in the vertical direction, these forces are equal and opposite so the normal force is equal to the 35 lb weight of the crate. The frictional force is therefore f = .30 * 35 lb = 10.5 lb.
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05:16:38 gen phy 4.55 18 kg box down 37 deg incline from rest, accel .27 m/s^2. what is the friction force and the coefficient of friction?
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05:16:39 GOOD STUDENT SOLUTION: (I don't know why, but I was hoping you would pick an odd numbered problem here)Here goes.....For an 18kg box on an incline of 37 degrees with an acceleration of .270 m/s/s, I first drew out a diagram showing the forces involved. Next the forces had to be derived. First, to find the force associated with the weight component parrallel to the inline moving the box downward....Fp=sin 37 deg(18kg)(9.8m/s/s)=106N. Next, the Normal force that is counter acting the mg of the box is found by.. Fn=cos 37 deg. (18kg)(9.8 m/s/s) = 141N. The frictional force can be found by using F=(mass)(acceleration) where (Net Force)-(frictional coeffecient*Normal Force)=(m)(a) so that... 106N - (141N * Friction Coeff.) = (18kg)(.270 m/s/s) where by rearranging, the frictional coeffecient is seen to be .717. INSTRUCTOR COMMENT: Good solution. Note that you should specify an x axis oriented down the incline, so that the acceleration will be positive. The weight vector being vertical in the downward direction is therefore in the fourth quadrant, at an angle of 37 degrees with respect to the negative y axis. Thus the weight vector makes angle 270 deg + 37 deg = 307 deg with the positive x axis and its x and y components are wtx = 18 kg * 9.8 m/s^2 * cos(307 deg) = 106 N and wty = 18 kg * 9.8 m/s^2 * sin(307 deg) = -141 N. You get the same results using the sin and cos of the 37 deg angle. The only other y force is the normal force and since the mass does not accelerate in the y direction we have normal force + (-141 N) = 0, which tells us that the normal force is 141 N. This also agrees with your result. **
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05:23:52 Univ. 5.90 (5.86 10th edition). 4 kg and 8 kg blocks, 30 deg plane, coeff .25 and .35 resp. Connected by string. Accel of each, tension in string. What if reversed?
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05:25:02 ** We will use the direction down the incline as the positive direction in all the following: The normal forces on the two blocks are 4 kg * 9.8 m/s^2 * cos(30 deg) = 34 N, approx., and 8 kg * 9.8 m/s^2 * cos(30 deg) = 68 N, approx. If sliding the 4 kg block will therefore experience frictional resistance .25 * 34 N = 8.5 N, approx. and the 8 kg block a frictional resistance .35 * 68 N = 24 N, approx. The gravitational components down the incline are 4 kg * 9.8 m/s^2 * sin(30 deg) = 19.6 N and 8 kg * 9.8 m/s^2 * sin(30 deg) = 39.2 N. If the blocks were separate the 4 kg block would experience net force 19.6 N - 8.5 N = 11.1 N down the incline, and the 8 kg block a net force of 39.2 N - 24 N = 15.2 N down the incline. The accelerations would be 11.1 N / (4 kg) = 2.8 m/s^2, approx., and 15.2 N / (8 kg) = 1.9 m/s^2, approx. If the 4 kg block is higher on the incline than the 8 kg block then the 4 kg block will tend to accelerate faster than the 8 kg block and the string will be unable to resist this tendency, so the blocks will have the indicated accelerations (at least until they collide). If the 4 kg block is lower on the incline than the 8 kg block it will tend to accelerate away from the block but the string will restrain it, and the two blocks will move as a system with total mass 12 kg and net force 15.2 N + 11.1 N = 26.3 N down the incline. The acceleration of the system will therefore be 26.3 N / (12 kg) = 2.2 m/s^2, approx.. In this case the net force on the 8 kg block will be 8 kg * 2.2 m/s^2 = 17.6 N, approx.. This net force is the sum of the tension T, the gravitational component m g sin(theta) down the incline and the frictional resistance mu * N: Fnet = T + m g sin(theta) - mu * N so that T = Fnet - m g sin(theta) + mu * N = 17.6 N - 39.2 N + 24 N = 2.4 N approx., or about 2.4 N directed down the incline. The relationship for the 4 kg mass, noting that for this mass T 'pulls' back up the incline, is Fnet = m g sin(theta) - T - mu * N so that T = -Fnet + m g sin(theta) - mu * N = -8.8 N + 19.6 N - 8.5 N = -2.3 N. equal within the accuracy of the mental approximations used here to the result obtained by considering the 8 kg block and confirming that calculation. **
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TWVபv닣ebw assignment #003 骶͌Άrzvc֡ Liberal Arts Mathematics I 08-04-2008
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05:31:04 Practice Test 2 Problem Number 1. Explain how we used a rubber band and a rail to demonstrate the conservation of the F `ds quantity.
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05:31:14 GOOD STUDENT ANSWER: We calculate the potential energy of the system when the rubber band is fully stretched, and compare with the F `ds total as the rail slides across the floor to see if all the potential energy was dissipated against friction.
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05:31:44 Problem Number 2 An Atwood machine consists of 34 paper clips, each of mass .4 grams, suspended from each side of a light pulley. If 4 clips are transferred from one side to the other, what will be the total gravitational force on the system?
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05:32:20 GOOD STUDENT SOLUTION: Well Im assuming that there are 34 paper clips on each side. So; 34 * .0004 kg= .0136 kg on one side when even 4 * .0004 kg = .0016 kg less on the side that has 4 removed so that gives you .0152 kg on one side and .012 kg on the other now to get the gravitational force on the system you multiply both by 9.8 m/sec^2 and that will give you the force acting on each one, which are: .14896 N and .1176 N respectively. Now to get the net gravitational force just subtract and that will give you a net gravitational force of GravFnet = .03136 N
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05:32:31 If the frictional force exerted by the pulley is .05 times the total weight of the system, then what is the net accelerating force?
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05:32:50 well take the weights given above and combine them, then multiply by the .05, that will give us: [(.0136 kg + .0016 kg) * 9.8 m/sec^2] * .05= .007448 N for the frictional force now subtract GravFnet Ff = Fnet, or 0.03136 N - .007448 N = .023912 N net force.
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05:32:53 What therefore is the acceleration of the system?
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05:32:57 Divide net force by the total mass of the system to get the acceleration of the system .023912 N / .0152 kg = 1.573 m/sec^2
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05:33:08 Problem Number 3 What will be the tension in the string holding a ball which is being swung in a circle of radius 1 meters, if the ball is making a complete revolution every .3 seconds? Assume that the system is in free fall (e.g., in a freely falling elevator, in orbit, etc.)?
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05:33:41 STUDENT ATTEMPT: Alright Im not very good at this YET, so I could use some help!!!!!!!! I know what is given in the problem. From this I can figure: angular velocity = 2pi*(f) where f = 3.33333 rev/sec so angular velocity = 20. 94359102 sec^-1 and because 1m* omega = linear velocity is 20. 944 m/sec AND THE REST I GET LOST ON, IM HAVING PROBLEMS WITH ALL THESE RELATIONSHIPS!!!!!!!!!!!!!!!! INSTRUCTOR NOTE: ** Centripetal acceleration is v^2 / r. That's the key thing you're forgetting here. This gives you aCent = (20.94 m/s)^2 / (1 m)^2 = 440 m/s^2. The centripetal force would therefore be mass_ball * 440 m/s^2. **
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05:33:48 What would be the tension in the string if the system was on and stationary with respect to the surface of the Earth, with the ball being swung in a vertical circle, when the ball is at the top of its arc?
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05:34:13 The net force on the ball is always equal to the centripetal force Fcent = mass_ball * 440 m/s^2. Gravity exerts a force equal to mass_ball * 9.8 m/s^2. If the ball is at the top of its arc the net force consists of the tension plus the gravitational force, both of which act in the same direction, which is downward. So using downward as the positive direction we have tension + mass_ball * 9.8 m/s^2 = mass_ball * 440 m/s^2. We find that tension = mass_ball * 440 m/s^2 - mass_ball * 9.8 m/s^2 = mass_ball * 430.2 m/s^2. You could substitute a reasonable mass (e.g., .4 kg) for the ball and obtain all quantities in Newtons. Note that these calculations are not in fact accurate to 4 significant figures. The numbers shown here are intended to demonstrate how the tension differs from the centripetal force.
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05:34:15 What if the ball is at the bottom of its arc?
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05:34:18 If the ball is at the bottom of its arc the net force consists of the tension plus the gravitational force, with tension up and gravity down. Again using downward as the positive direction and noting that the upward centripetal acceleration is with this assumption negative we have tension + mass_ball * 9.8 m/s^2 = -mass_ball * 440 m/s^2. We find that tension = -mass_ball * 440 m/s^2 - mass_ball * 9.8 m/s^2 = -mass_ball * 449.8 m/s^2.
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05:34:20 What if the ball is at its halfway height?
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05:34:28 At the halfway height the centripetal force and gravitational acceleration are perpendicular and hence independent. The only force acting toward the center is the tension in the string, which must therefore supply the entire centripetal force. The tension is mass_ball * 440 m/s^2 toward the center.
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08:20:45 Problem Number 4 If we set the expression G M / r^2 for the gravitational acceleration at distance r from the center of a planet of mass M equal to the centripetal acceleration v^2 / r and solve for v in terms of r, what is the result?
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08:20:55 Setting the two expressions equal we have the equation G M / r^2 = v^2 / r. To solve for v we first multiply both sides by r to get G M r = v^2, Taking the square root of both sides and reversing sides of the equation gives us v = sqrt(G M r).
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08:21:09 Problem Number 5 A disk of negligible mass and radius 30 cm is constrained to rotate on a frictionless axis about its center. On the disk are mounted masses of 7 gram at a distance of 23.1 cm from the center, 24 grams data distance of 16.8 cm from the center and 47 grams at a distance o 9 cm from the center. A uniform force of .06983 Newtons is applied at the rim of the disk in a direction tangent to the disk.
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08:21:14 What will be the angular acceleration of the disk?
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08:21:28 ** The moment of inertia of the system is the sum of all the m r^2 contributions of the individual particles. The net torque is the product of the net force and the moment arm. Newton's 2d law F = m a, expressed in in angular form, is `tau = I `alpha, where `tau is net torque (analogous to force), I is moment of intertia (sum of all mr^2, analogous to mass) and `alpha is angular acceleration in rad/s^2. **
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08:21:34 If the force is applied for 4 seconds with the disk initially at rest, what angular velocity with the disk attain?
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