course PHY201 PNiβJ~wÏZLwassignment #020
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16:55:15 `q001. Note that this assignment contains 3 questions. . A 5 kg block rests on a tabletop. A string runs horizontally from the block over a pulley of negligible mass and with negligible friction at the edge of the table. There is a 2 kg block hanging from the string. If there is no friction between the block in the tabletop, what should be the acceleration of the system after its release?
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RESPONSE --> Gravitational force on 2 kg block: 2 kg * 9.8 m/s^2 = 19.6 N Acceleration of 5 kg block: 19.6 N / 7 kg = 2.8 m/s^2 confidence assessment: 2
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16:55:50 Gravity exerts a force of 5 kg * 9.8 meters/second = 49 Newtons on the block, but presumably the tabletop is strong enough to support the block and so exerts exactly enough force, 49 Newtons upward, to support the block. The total of this supporting force and the gravitational force is zero. The gravitational force of 2 kg * 9.8 meters/second = 19.6 Newtons is not balanced by any force acting on the two mass system, so we have a system of total mass 7 kg subject to a net force of 19.6 Newtons. The acceleration of this system will therefore be 19.6 Newtons/(7 kg) = 2.8 meters/second ^ 2.
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RESPONSE --> self critique assessment: 3
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16:59:41 `q002. Answer the same question as that of the previous problem, except this time take into account friction between the block in the tabletop, which exerts a force opposed to motion which is .10 times as great as the force between the tabletop and the block. Assume that the system slides in the direction in which it is accelerated by gravity.
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RESPONSE --> Gravitational force on 2 kg block: 2 kg * 9.8 m/s^2 = 19.6 N Gravitational force on 5 kg block: 5 kg * 9.8 m/s^2 = 49 N Friction on 5 kg block: .1 * 49 N = 4.9 N Acceleration of 5 kg block: (19.6 N - 4.9 N) / 7 kg = 2.1 m/s^2 confidence assessment: 2
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16:59:49 Again the weight of the object is exactly balance by the upward force of the table on the block. This force has a magnitude of 49 Newtons. Thus friction exerts a force of .10 * 49 Newtons = 4.9 Newtons. This force will act in the direction opposite that of the motion of the system. It will therefore be opposed to the 19.6 Newton force exerted by gravity on the 2 kg object. The net force on the system is therefore 19.6 Newtons -4.9 Newtons = 14.7 Newtons. The system will therefore accelerate at rate a = 14.7 Newtons/(7 kg) = 2.1 meters/second ^ 2.
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RESPONSE --> self critique assessment: 3
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17:16:07 `q003. Answer the same question as that of the preceding problem, but this time assume that the 5 kg object is not on a level tabletop but on an incline at an angle of 12 degrees, and with the incline descending in the direction of the pulley.
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RESPONSE --> Gravitational force on 2 kg block: 2 kg * 9.8 m/s^2 = 19.6 N Gravitational force on 5 kg block: 5 kg * 9.8 m/s^2 = 49 N Parallel component of force on 5 kg block: 49 N * sin(12) = 10.19 N Friction on 5 kg block: .1 * (49 N - 10.19 N) = 3.9 N Acceleration of 5 kg block: (19.6 N + 10.19 N - 3.9 N) / 7 kg = 3.7 m/s^2 confidence assessment: 2
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17:30:56 In this case you should have drawn the incline with the x axis pointing down the incline and the y axis perpendicular to the incline. Thus the x axis is directed 12 degrees below horizontal. As a result the weight vector, rather than being directed along the negative y axis, lies in the fourth quadrant of the coordinate system at an angle of 12 degrees with the negative y axis. So the weight vector makes an angle of 270 degrees + 12 degrees = 282 degrees with the positive x axis. The weight vector, which has magnitude 5 kg * 9.8 meters/second ^ 2 = 49 Newtons, therefore has x component 49 Newtons * cosine (282 degrees) = 10 Newtons approximately. Its y component is 49 Newtons * sine (282 degrees) = -48 Newtons, approximately. The incline exerts sufficient force that the net y component of the force on the block is zero. The incline therefore exerts a force of + 48 Newtons. Friction exerts a force which is .10 of this force, or .10 * 48 Newtons = 4.8 Newtons opposed to the direction of motion. Assuming that the direction of motion is down the incline, frictional force is therefore -4.8 Newtons in the x direction. The weight component in the x direction and the frictional force in this direction therefore total 10 Newtons + (-4.8 Newtons) = + 5.2 Newtons. This force tends to accelerate the system in the same direction as does the weight of the 2 kg mass. This results in a net force of 5.2 Newtons + 19.6 Newtons = 24.8 Newtons on the 7 kg system. The system therefore accelerates at rate {} a = (24.8 Newtons) / (7 kg) = 3.5 meters/second ^ 2, approximately.
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RESPONSE --> I mistakenly assumed that the x component of the force would be subtracted from the y component. It doesn't make much sense to me that the component forces could sum to a force that is greater than the original force so I simply subtracted the one component from the total. Looking back this was naive. Also, I did my calculations based on sin(12) and cos(12) which gives a similar ratio but flips the order of the values. I need to think more about the orientation. Is there a difference between using the tilt of the ramp as 258 degrees or 282 degrees as far as you are concerned? The results of sin and cos are the same for those values.
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