Constraint equations in mechanics

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Kushal Chakrabarti
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1. The problem: Two blocks of mass m=5kg and M= 10kg are connected by a string passing over a pulley B. Another string connects pulley B to the floor and passes over pulley A. An upward force F is applied at the centre of pulley A. Both pulleys are massless. Find acceleration of the blocks if F is:
a) 100N, b) 300N, c) 500N
Take g=10 ms-2
I have attached an image of the problem.

Attempt at a solution:
Well I drew an FBD in the frame of pulley A, thus using a downward pseudo force. Also, I found using constraint equations that F=4T, where T is the tension in the thread joining the two masses. Thus it gave me two equations ( one for each mass) but solving them, I found out that both the accelerations are turning out to be negative. Where did I go wrong?
Diagram of a pulley system: force F pulls upward on the centre of pulley A, which hangs from a string running down to the floor; pulley B hangs below A on a string, with blocks m and M on either end of a string over B resting on the floor.
 
Attachments
  • Diagram of a pulley system: force F pulls upward on the centre of pulley A, which hangs from a string running down to the floor; pulley B hangs below A on a string, with blocks m and M on either end of a string over B resting on the floor.
    15320104568961692308584.jpg
    24 KB · Views: 813
Kushal Chakrabarti said:
Thus it gave me two equations ( one for each mass) but solving them, I found out that both the accelerations are turning out to be negative. Where did I go wrong?
How are we going to have any chance to answer that if you do not show us your calculations?
 
Orodruin said:
How are we going to have any chance to answer that if you do not show us your calculations?
I'm very sorry sir I'll put my attempt too
 
Handwritten working on lined paper: a small pulley diagram with force F and tension T labels above free-body sketches of blocks m and M with pseudo force terms, followed by equations T - (mg + ma) = ma1 and T - (Mg + Ma) = Ma2 and results a1 = -35/3 m/s^2 and a2 = -20/3 m/s^2.
 
Attachments
  • Handwritten working on lined paper: a small pulley diagram with force F and tension T labels above free-body sketches of blocks m and M with pseudo force terms, followed by equations T - (mg + ma) = ma1 and T - (Mg + Ma) = Ma2 and results a1 = -35/3 m/s^2 and a2 = -20/3 m/s^2.
    15320118790831916162559.jpg
    39.5 KB · Views: 1,232
Orodruin said:
How are we going to have any chance to answer that if you do not show us your calculations?
I've now added my attempt too, sir
 
Kushal Chakrabarti said:
Please take the trouble to type working. Images are for diagrams and textbook extracts.
If you must post as images, please number equations for reference.

The line F=(M+m)a omits that a cord is anchored to the floor. It is also wrong in supposing that the mass centre of the two masses accelerates at the same rate as the upper pulley.