course Phy231 5/11/10 3:30pm 20.Q&A020. Forces (inclines, friction)
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*ent:ok ********************************************* Question: `q002. Answer the same question as that of the previous problem, except this time take into account friction between the block in the tabletop, which exerts a force opposed to motion which is .10 times as great as the force between the tabletop and the block. Assume that the system slides in the direction in which it is accelerated by gravity. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F=5kg*9.8m/s^2=49N Ffrict=49N*.1=4.9N This will be in the opposite direction of the force of the system by gravity. Fnet=19.6N-4.9N=14.7N acceleration=14.7N/7kg=2.1m/s^2 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&*ent:ok ********************************************* Question: `q003. Answer the same question as that of the preceding problem, but this time assume that the 5 kg object is not on a level tabletop but on an incline at an angle of 12 degrees, and with the incline descending in the direction of the pulley. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x-component=49N*cos(282deg)=10.2N y-component=49N*sin(282deg)=-47.9N The Ffrict is going to be in the opposite direction of the x-axis. x-component=10.2N-4.8N=5.4N This will be going in the same direction as the 2kg force. Fnet=5.4N+19.6N=25N acceleration=25N/7kg=3.57m/s^2 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. STUDENT COMMENT: i get confused every time about which angle to use and how to determine it, where did 270 deg come from, i used the right equations and was on to the right answer but used the wrong numbers in the equations, but i had the right idea ------------------------------------------------ Self-critique rating #$&*ent: 1 INSTRUCTOR RESPONSE: The negative y axis lies at 270 deg from the positive x axis, as measured counterclockwise. When the coordinate system is rotated 12 degrees in the manner described, the weight vector stays where it is, but the y negative axis swings 'back' 12 degrees and the angle of the weight vector becomes 282 degrees. MORE EXTENSIVE EXPLANATION The weight is in the downward vertical direction, which matches the direction of the original vertical-horizontal coordinate system. So in the original system the weight vector is at 270 degrees. However the positive x axis of the original coordinate system doesn't match the direction of motion along the incline. It's generally simpler to have the x axis parallel to the direction of motion. To accomplish this we rotate the coordinate system 12 degrees in the clockwise direction. As the coordinate system is rotated, the positive x axis rotates to an orientation 12 degrees below horizontal, and the negative y axis 'swings out' 12 degrees from its original vertical orientation. This leaves the vertical weight vector in the fourth quadrant, 12 degrees from the newly oriented negative y axis. As measure counterclockwise from the positive x axis, the weight vector is now at angle 282 degrees. The first figure below depicts a weight vector with its initial point at the origin of an x-y coordinate system in standard vertical-horizontal orientation. It should be clear that the vector is at angle 270 degrees, as measured counterclockwise from the positive x axis. The second figure shows the same weight vector, which is still vertical, but with the coordinate system rotated so that the positive x axis is directed down a 12-degree incline. It should be clear that negative y axis will have rotated 12 degrees from its original position so that the weight vector is now at 282 degrees. STUDENT QUESTION I have read both of the explanatons I am still not sure where the 270 degrees is coming from, how do we calculate this INSTRUCTOR RESPONSE The above discussion is fairly long, and involves an idea that is difficult for many students. The following should provide a convenient summary of the key ideas. If this point-by-point summary doesn't clarify everything for you, please submit a copy of these questions along with inserted comments, questions, etc. (insertions should be marked with &&&&). 1.If the x-y coordinate system is in its 'standard orientation', with the x axis horizontal, pointing to the right, and the y axis vertical and upward, then gravity pulls straight down, along the negative y axis. 2.The negative y axis of this system is at 270 degrees as measured counterclockwise from the positive x axis. 3.If the coordinate system is rotated 12 degrees in the clockwise direction, then the x axis now points slightly below horizontal, and the y axis is 12 degrees from vertical. 4.The gravitational force is still vertical, so the gravitational force is no longer in the direction of the negative y axis. It is now directed at an angle of 12 degrees from the negative y axis. 5.The gravitational force is no longer at 270 degrees, since the rotation of the system has rotated the y axis away from vertical. 6.The gravitational force is now in the fourth quadrant of the rotated system, directed at 282 degrees. #$&* "