4. (15 pts) Newton's Law In order that two boxes, one on...

4. (15 pts) Newton's Law In order that two boxes, one on top of the other, are sliding down the ramp, together with a same constant speed, a force F is applied to the box B in the opposite direction of the motion, as shown in figure. The coefficient of static friction, ??s = 0.90 is between the two boxes. The coefficient of the kinetic friction between the box B and the ramp, is ??k = 0.50 . (Take g = 9,8 m/s^2; m_A=1 kg; m_B=9 kg; sin 30^0=0.5, cos 30^0= 0.87) a) Draw free body diagrams of the mass A and B b) Write the equations of the motions of the mass A and B c) Find the force F if the angle ?? = 37^0 d) Find the static friction force if the angle ?? = 30^0 e) Find the maximum value of ?? such that the mass A does not move with respect to B.

Transcript

00:01 In the given problem we have two boxes a and b that are kept one top of the other, that is a is kept on top of box b and they are sliding down the wrap together with the same constant speed.

00:11 A force f is applied to box b that moves in the opposite direction.

00:17 The coefficient of static friction given to us is 0 .9, while the coefficient of kinetic friction is 0 .5.

00:23 We have that mass of box a is equal to 1kg.

00:30 Mass of box v is 9 kg and we are given the values of sine 30 and cos 30 degree in part a of the problem we have to first draw the free body diagrams of the both the masses so when we see that this mass is kept on the another mass so for a the process acting will be first the frictional force which is in this direction let this be the y -axis and this with the x -axis so this will be towards the x -axis which is along the ramp the weight of the box will be vertically downward which is given by m -a -g and the normal force acting due to box b will be n1 so resolving this weight into two components along the x and the y -axis we can get this is the angle alpha hence this horizontal component or the vertical component will be alpha and the force along the x -axis will be m a g sine alpha this is the free body diagram of box e now let us look at the free body diagram of box v hence if this is the incline and this is the box v the normal force due to the incline will be in the upward direction let it be denoted by n2 and the normal force due to n due to box 1 will be in the downward direction and this is denoted by n1 the weight of the box will be vertically downward denoted by m b g the there is a force f acting along the incline in the opposite direction and the frictional force here will be f2 while the frictional force due to box one here will be f1 again resolving the weight we will get m b g cos theta in the negative y direction and here n x m b g sine theta in the positive y direction but this m a mb will be the sum of both the boxes that is m a plus mb so this is the free body diagram of box b now in part b we have to write the equations of motions of both the boxes for a we can write that f1 is equal to m a g sign alpha on resolving the x axis and for on resolving the y -axis we can write that n1 is equal to m a g -gos alpha this both is for box a now for box b we can write the two forces along the x -axis it will be f1 plus m b g -sign alpha this is equal to f plus f2 this will be our equation three and along the y -a axis we can write that if n2 is equal to m b g cos alpha this will be our fourth equation so these two equations are for box b now in part c we have to find the force f if the angle alpha given to us is 37 degrees so we have equation one and equation four equation one which will give us m a g sign we will substitute m a g sine theta into this equation three here and we will calculate the force f so this will be m a g sine alpha plus m b g sign alpha which will be be equal to f plus the other force here is f2 that is frictional force 2 so this will be written as mu k and this the the normal force here is m a plus m mb g, cos alpha.

05:20 We have all the values here except for f.

05:23 So we will calculate f as ma plus mb times g times sine alpha minus mu k g g cos alpha.

05:40 Substituting the values now.

05:43 Ma is 1 kg.

05:45 Mb is 9 kg.

05:47 Acceleration due to gravity is 9 .8 meter per second square.

05:52 Sine of 37 degrees minus mu k is 0 .5 times 9 .8 meter per second square times cause alpha.

06:05 So this will be cause of 37 degrees...

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