Physics Numericals based on Friction

Question

Two bodies, A(100g) and B(500g), are connected by a pulley. The system rest horizontally. The pulley is connected to a wall while the bodies rest on a horizontal surface. A lies on top of B. What force F is required to give B an acceleration of 20cm/s^2?

(coefficient of friction between the two bodies & also between the surface and bodies is 0.2)

Answer 1

This problem can be solved by setting up two free body diagrams:  one for mass A, the other for mass B.

The cord connecting the two masses will have some tension T, which is the same everywhere in the cord.  Let the coefficient of friction be μ,the acceleration of gravity be g,  the mass of A   be m,  the mass of B  be M and the acceleration be a.   The magnitude of the acceleration of A must be the same as the magnitude of the acceleration of B - this magnitude is what the symbol a will represent.  Of course the two masses move in opposite directions:  A to the left and B to the right.  However, I will take the positive direction of A to be to the left and the positive direction of B to be to the right.  This choice is motivated by the standard approach to Atwood's machine.

 

  It is easier to set up the equations using symbols and then plugging in numerical values at the end.

 

The net force on A is   T - mμg , so Newton's second law reads   T- mμg =  m a   solving for T gives

       T =  ma + mμg

 

The net force on B  is   F - T - mμg - (m + M) μg ,  so Newton's second law reads

 

       F - T - mμg -(m + M) g =   M a    substituting for T and collecting terms gives

 

  F =  (M + m) a + (3 m +M) μg

 

  Note that both the force on A and the force on B have a contribution  -mμg.   This is a consequence of Newton's third law.

 

   Plugging in numbers (after converting to kg, meters and m/s/s) gives    F = .12 + .16 g  . 

Taking g = 9.81  gives       F = 1.69 N.