#$&* course PHY 201 7/1/2012 5:25 Samantha RogersPHY 201
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Given Solution: 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. STUDENT QUESTION since it is a frictionless system i thought the answer would the acceleration of gravity since that would be the force pulling the objects down. INSTRUCTOR RESPONSE A mass resting by itself on a level tabletop does now accelerate downward in response to the gravitational force exerted on it. This is because the tabletop pushes up on it. This occurs whether friction is high, low or absent. Of course if friction is very low, you have to be sure the tabletop is very nearly level, and if friction is absent you'd best be sure it is completely level. These forces are still present if you add the hanging mass to the system. The mass on the tabletop does experience the downward pull of gravity, but that force is balanced by the supporting force of the tabletop. The mass therefore does not move in the direction of the gravitational pull. It is nevertheless part of the system being accelerated by the gravitational force on the hanging mass. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The total force applied is still 19.6 N, but we will find the net forcing acting upon the system by subtracting the Ffriction = 5 kg * 9.8 m/s^2 * .1 = 4.9 N; therefore, the net force is Fnet = 19.6 N - 4.9 N = 14.7 N. This gives us an acceleration of a = 14.7 N/ 7 kg = 2.1 m/s^2 confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to find the net force we must subtract from our total force of 19.6 N the values of force caused by friction and gravity. The gravitational force is found by 5 kg * 9.8 m/s^2 * sin (12 degrees) = 10.2 N, and the frictional force is determined by the normal force acting upon 5 kg * 9.8 m/s^2 * cos (12 degrees) * .1 = 4.8 N; therefore, the net force of the system is 19.6 N - 4.8 N - 10.2 N = 4.6 N. This gives us an acceleration of 4.6 N/ 7 kg = 0.66 m/s^2 = a. confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to find the net force we must subtract from our total force of 19.6 N the values of force caused by friction and gravity. The gravitational force is found by 5 kg * 9.8 m/s^2 * sin (12 degrees) = 10.2 N, and the frictional force is determined by the normal force acting upon 5 kg * 9.8 m/s^2 * cos (12 degrees) * .1 = 4.8 N; therefore, the net force of the system is 19.6 N - 4.8 N - 10.2 N = 4.6 N. This gives us an acceleration of 4.6 N/ 7 kg = 0.66 m/s^2 = a. confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to find the net force we must subtract from our total force of 19.6 N the values of force caused by friction and gravity. The gravitational force is found by 5 kg * 9.8 m/s^2 * sin (12 degrees) = 10.2 N, and the frictional force is determined by the normal force acting upon 5 kg * 9.8 m/s^2 * cos (12 degrees) * .1 = 4.8 N; therefore, the net force of the system is 19.6 N - 4.8 N - 10.2 N = 4.6 N. This gives us an acceleration of 4.6 N/ 7 kg = 0.66 m/s^2 = a. confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!