course Phy 121
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19:48:02 `q001. Note that this assignment contains 5 questions. . A pendulum consists of a 150 g mass suspended from a light string. Another light string is attached to the mass, which is then pulled back from its equilibrium position by that string until the first string makes an angle of 15 degrees with vertical. The second string remains horizontal. Let the x axis be horizontal and the y axis vertical. Assume that the mass is pulled in the positive x direction. If T stands for the tension in the pendulum string, then in terms of the variable T what are the x and y components of the tension?
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RESPONSE --> m = 150g = 0.15kg Since the object is pulled towards the positive x-axis, the first string will make a 15 degree angle to the left of the vertical y-axis. Therefore the angle of tension from the x-axis is 90 degrees more. theta = 15 degrees + 90 degrees = 105 degrees We do not know the force of tension, but we can make our equation for the x and y components using the direction of that force. Tx = T * cos(105) Ty = T * sin(105)
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19:48:16 The pendulum string makes an angle of 15 degrees with vertical. Since we have assumed that the pendulum is pulled in the positive x direction, the direction of the tension in the string will be upward and to the left at an angle of 15 degrees with vertical. The tension force will therefore be directed at 90 degrees + 15 degrees = 105 degrees as measured counterclockwise from the positive x axis. The tension will therefore have x component T cos(105 degrees) and y component T sin(105 degrees).
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RESPONSE --> I understand.
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19:52:42 `q002. Continuing the preceding problem, we see that we have a vertical force of T sin(105 deg) from the tension. What other vertical force(s) act on the mass? What is the magnitude and direction of each of these forces? What therefore must be the magnitude of T sin(105 deg).
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RESPONSE --> Gravity is the vertical force that acts on the object. To find the magnitude of gravitational force, we multiply the mass of the object by the acceleration of gravity. F = m*g = (0.15kg)(9.8m/s^2) = 1.47 Newtons. Since this is the only vertical force, the magnitude of T sin(105 deg) is equal to the force of gravity. T*sin(105deg) = 1.47 N
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19:54:15 The only other vertical force acting on the mass will be the gravitational force, which is .150 kg * 9.8 meters/second ^ 2 = 1.47 Newtons. The direction of this force is vertically downward. Since the mass is in equilibrium, i.e., not accelerating, the net force in the y direction must be zero. Thus T sin(105 deg) - 1.47 Newtons = 0 and T sin(105 deg) = 1.47 Newtons.
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RESPONSE --> I understand this problem.
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19:58:10 `q003. Continuing the preceding two problems, what therefore must be the tension T, and how much tension is there in the horizontal string which is holding the pendulum back?
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RESPONSE --> Since we know the other elements of the function, we can now find the value for tension. T * sin(105) = 1.47N T = (1.47N) / sin(105) = 1.52 Newtons Now that we know the force of tension, we can use it to find the horizontal component. Tx = (1.52N)(sin 105deg) = -0.39N Now in order to find the horizontal force, we subtract the x-component from the net force on the second string. The object is not moving, and this means there is a balance of forces applied to the object. Therefore, the net force is zero. 0 - -0.39N = 0.39N
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19:58:53 If T sin(105 deg) = 1.47 Newtons then T = 1.47 Newtons / (sin(15 deg)) = 1.47 Newtons/.97 = 1.52 Newtons. Thus the horizontal component of the tension will be T cos(105 deg) = 1.52 Newtons * cos(105 deg) = 1.52 Newtons * (-.26) = -.39 Newtons, approximately. Since the mass is in equilibrium, the net force in the x direction must be zero. The only forces acting in the x direction are the x component of the tension, to which we just found to be -.39 Newtons, and the tension in the second string, which for the moment will call T2. Thus T2 + (-.39 N) = 0 and T2 = .39 N. That is, the tension in the second string is .39 Newtons. STUDENT COMMENT: I'm really confused now. If we started out with a .15 kg mass that is equal to 1.47 Newtons. How did we create more weight to get 1.52 Newtons? Is the horizontal string not helping support the weight or is it puling on the weight adding more force? INSTRUCTOR RESPONSE: A horizontal force has no vertical component and cannot help to support an object against a vertical force. The vertical component of the tension is what supports the weight, so the tension has to be a bit greater than the weight. The tension in the string is resisting the downward weight vector as well as the horizontal pull, so by the Pythagorean Theorem it must be greater than either.
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RESPONSE --> I understand this problem.
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20:02:25 `q004. If a 2 kg pendulum is held back at an angle of 20 degrees from vertical by a horizontal force, what is the magnitude of that horizontal force?
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RESPONSE --> m = 2kg If the mass is 20 degrees from vertical it is 110 degrees from the x-axis. (20 deg + 90 deg = 110 deg) The magnitude of the vertical force can be found by multiplying the mass by the vertical acceleration. (F = 2kg * 9.8m/s^2 = 19.6 N). Now we can solve for Ty knowing that Ty = T * sin(110 deg) = 19.6N T = (19.6N)/sin(110deg) = 20.86 N And now, solve for the horizontal component: Tx = 20.86N * cos(110 deg) = -7.13 Newtons. The horizontal force is equal and opposite to the horizontal component for an object in equilibrium. Therfore, Fx = 7.13N.
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20:04:13 At the 20 degree angle the tension in the pendulum string will have a vertical component equal and opposite to the force exerted by gravity. The tension with therefore have a horizontal component. To achieve equilibrium by exerting the horizontal force, this horizontal force must balance the horizontal component of the tension. We therefore begin by letting T stand for the tension in the pendulum string. We also assumed that the pendulum is displaced in the positive x, so that the direction of the string as measured counterclockwise from the positive x axis will be 90 degrees + 20 degrees = 110 degrees. Thus the x component of the tension will be T cos(110 deg) and the y component of the tension will be T sin(110 deg). The weight of the 2 kg pendulum is 2 kg * 9.8 meters/second ^ 2 = 19.6 Newtons, directed in the negative vertical direction. Since the pendulum are in equilibrium, the net vertical force is zero: T sin(110 deg) + (-19.6 N) = 0 This equation is easily solved for the tension: T = 19.6 N / (sin(110 deg) ) = 19.6 N / (.94) = 20.8 Newtons, approximately. The horizontal component of the tension is therefore T cos(110 deg) = 20.8 N * cos(110 deg) = 20.8 N * (-.34) = -7 N, approx.. To achieve equilibrium, the additional horizontal force needed will be + 7 Newtons.
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RESPONSE --> Ok
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20:13:16 `q005. The 2 kg pendulum in the previous exercise is again pulled back to an angle of 20 degrees with vertical. This time it is held in that position by a chain of negligible mass which makes an angle of 40 degrees above horizontal. Describe your sketch of the forces acting on the mass of the pendulum. What must be the tension in the chain?
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RESPONSE --> My sketch of these forces shows one vector representing the weight of the object in the vertical downward direction. Then one line, representing the first string, is 20 degrees to the left of the positive y-axis. Then the chain is represented by a line that is 40 degrees above the positive x-axis. m = 2kg theta 1 = 20 deg + 90 deg = 110 deg theta 2 = 40 deg In this case, we have two angles and two sets of components which make two tension vectors. So we will have to add the two vectors together. Tx = T1 * cos(110) + T2 * cos(40) Ty = T1 * sin(110) + T2 * sin(40) We also know from the problem that the chain's gravitational force is zero because the chain is horizontal and the object is in equilibrium. The gravitational force on the string is 19.6N downward. Now we can combine our information to solve for tension. Tx = T1 * cos(110) + T2 * cos(40) + 0 = (-0.34 T1) + (0.766 T2) T1 = (-0.766/-0.34) T2 = 2.25 T2 Ty = T1 * sin(110) + T2 * sin(40) - 19.6N (0.94 T1) + (0.643 T2) = 19.6N (0.94)(2.25 T2) + 0.7643 T2 = 19.6N 2.88 T2 = 19.6N T2 = 6.8 N T1 = 2.25 T2 = (2.25)(6.8N) = 15.3N
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20:18:48 The weight of the pendulum is partially supported by the tension in the chain. Thus the tension in the pendulum string is not the same as before. The horizontal component of the tension in the chain will be equal and opposite to the horizontal component of the tension in the pendulum string. Your picture should show the weight vector acting straight downward, the tension in the pendulum string acting upward and to the left at an angle of 20 degrees to vertical and the tension in the chain should act upward into the right at an angle of 40 degrees above horizontal. The lengths of the vectors should be adjusted so that the horizontal components of the two tensions are equal and opposite, and so that the sum of the vertical components of the two tensions is equal of opposite to the weight vector. Since both tensions are unknown we will let T1 stand for the tension in the pendulum and T2 for the tension in the chain. Then T1, as in the preceding problem, acts at an angle of 110 degrees as measured counterclockwise from the positive x axis, and T2 acts at an angle of 40 degrees. At this point whether or not we know where we are going, we should realize that we need to break everything into x and y components. It is advisable to put this information into a table something like the following: x comp y comp T1 T1 * cos(110 deg) T1 * sin(110 deg) in T2 T2 * cos(40 deg) T2 * sin(40 deg) Weight 0 -19.6 N The pendulum is held in equilibrium, so the sum of all the x components must be 0, as must the sum of all y components. We thus obtain the two equations T1 * cos(110 deg) + T2 * cos(40 deg) = 0 and T1 * sin(110 deg) + T2 * sin(40 deg) - 19.6 N = 0. The values of the sines and cosines can be substituted into the equations obtain the equations -.33 T1 + .77 T2 = 0 .95 T1 + .64 T2 - 19.6 N = 0. We solve these two simultaneous equations for T1 and T2 using one of the usual methods. Here we will solve using the method of substitution. If we solve the first equation for T1 in terms of T2 we obtain T1 = .77 T2 / .33 = 2.3 T2. Substituting 2.3 T2 for T1 in the second equation we obtain .95 * 2.3 T2 + .64 T2 - 19.6 N = 0, which we easily rearrange to obtain 2.18 T2 + .64 T2 = 19.6 Newtons, or 2.82 T2 = 19.6 N, which has solution T2 = 19.6 Newtons/2.82 = 6.9 N, approximately. Since T1 = 2.3 T2, we have T1 = 2.3 * 6.9 N = 15.9 N, approximately. Thus the pendulum string has tension approximately 15.9 Newtons and the chain the tension of approximately 6.9 Newtons.
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RESPONSE --> Ok. My numbers were a little different but simply because of rounding.
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?I?P???????`??? assignment #025 025. `query 25 Physics I 07-12-2006
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12:40:13 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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RESPONSE --> Forces exerted on a baseball at the point of contact include gravitational force, the force behind the bat which is the normal force, and frictional force between the bat and the ball.
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12:42:06 ** 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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RESPONSE --> Ok. Is my response adequate?
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12:44:45 gen phy list the forces on the ball while flying toward the outfield, and describe the directions of these forces
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RESPONSE --> The forces exerted on the ball as it travels through air include gravitational force or weight, obviously, and air resistance. Gravity is a conservative force while air resistance, like friction, is a nonconservative force.
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12:47:26 **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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RESPONSE --> I understand.
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12:47:36 gen phy give the source of each force you have described
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RESPONSE --> The source of gravitational force is of course, gravity itself, the source of the normal force is the contact of the ball and the bat, friction is a result of the actual touching of the two surfaces at contact, and air resistance is obviously a result of the air exerting a force against the ball's direction of motion.
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12:47:51 ** 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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RESPONSE --> I did not explain that the normal force is a result of ""elastic compression.""
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12:48:41 gen phy what is the direction of the net force on the ball while in contact with the bat?
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RESPONSE --> The net force is illustrated by the direction of the ball. Therefore, the net force is forward and would have a downward concavity if graphed. The normal force is dominant immediately after contact, sending the ball into the air. The force of gravity is dominant in the end, pulling the ball downward.
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12:50:15 ** 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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RESPONSE --> I didn't know how to explain. Was my response wrong?
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12:51:10 ** 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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RESPONSE --> I understand.
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12:51:27 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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RESPONSE --> This problem does not apply to my course, and I do not know how to answer it.
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12:51:32 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 -->
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