Question

A painter of mass M stands on a platform of mass
pulley as shown. He pulls each rope with the force F and
stands on a platform of mass m and pulls himself up by two ropes which hang over
Find 'a'
He pulls each rope with the force F and moves upward with uniform acceleration 'a'.​

Answer 1

Answer:

Sum of the forces on the painter is  

F= 2T+N-Mg,                     (1)  

Where (T= tension in each rope, N is normal force of the platform on the painter.  

Sum of the forces on the platform is  

F= 2T-mg-N                        (2)

Where (N= painter exerts a force on the platform in the opposite direction and therefore their accelerations are also same ‘ma’

 

Explanation:

So Newton’s Second Law for each gives the following two equations:

2T+N-Mg=Ma                          (3)

2T-mg-N=ma                           (4)

There are three unknowns T, N, a, in these equations.  

But recall that the painter exerts a force F on each rope, and by Newton’s Third Law, the ropes exert a force F back on him, which is exactly the tension, so we have,  

T=F                                          (5)

Now we simply solve these three equations and three unknowns for a. By adding eq (3) + (4), which eliminates N and gives

4F-(M+m)g= (M+m)a               (6)

Solving for 'a' then gives  

a=4FM+m-g                              

Answer 2

The answer for your question is ,

a=( 4F÷(M+m)) - g