course Phy 201 Question: `q001. Note that this assignment contains 3 questions.
.............................................
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): Self-critique rating: ********************************************* 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 Weight of the block on the table top will again be balanced by the upward supporting force of the table. It will still have a magnitude of 49 N. To determine the frictional force we will have: .10 * 49 N = 4.9 N. The frictional force will act in the opposite direction of motion and will oppose the 19.6 N force that is exerted by gravity on the 2 kg hanging block. Fnet =19.6 N - 4.9 N = 14.7 N We can then calculate the new acceleration: a = 14.7 N/7 kg = 2.1 m/s^ 2 Confidence assessment rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): 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: X axis is in the direction 12 degrees below horizontal. Because of this, the vector for weight will be in the range of the 4th quadrant and will be at a 12 degree angle with that of the y axis which is negative. Therefore, the weight vector will make an angle with the positve x axis that = 270 deg + 12 deg = 282 deg The weight vector, having a magnitude = 5 kg * 9.8 m/s^ 2 = 49 N, will have an x component = 49 N * cos (282 deg) = approx 10 N It will have a y component = 49 N * sin (282 deg) = approx -48 N The incline will exert a force that will be enough so that the y component of the force acting on the block will be 0. Because of this we can determine that the incline is exerting a force of positive 48 N. Friction force .10 of the previous force and can be found by: .10 * 48 N = 4.8 N that are opposing the direction of motion. If we assume that direction of motion is downward along the incline, then Ffrict would have to be -4.8 in the direction of x. If the weight component and the frictional force are both in the direction of x, then there total = 10 N + (-4.8 N) = + 5.2 N This force will accelerate the system in the same direction as the weight of the 2 kg mass. So Fnet = 5.2 N + 19.6 N = 24.8 N acting on the 7 kg system. Acceleration will then be: a = 24.8 N / 7 kg = approx 3.5 m/s^2 Confidence assessment rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: 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. "