A question about masses on a pully

[babbagee]

Three block, connecting ropes, and a light frictionless pully comprise a system, as shown. An external force P is applied downward on block A. The system accelerates at the rate of 2.5m/s2. The tension in the rope connecting block B and block C equals 60 N. I don't have a picture but ill discribe it.

Ok there are two masses on one side of the pully B and C. B is on top and C is on bottom, the Force exerted on B by C is 60N. Mass of B is 18kg. Then there is a mass on the other side which has a mass of 12Kg.

Ok i tired this problem but i keep getting the wrong answer.

The total force on B is

T - (60 N + w_B) = m_aa

and then the total force on B is

T - w_A = -m_aa

Since A is accelerating downwards it has a negative acceleration and B has a positive acceleration.

The answer should be 190N but i can't seem to get the answer. What am i doing wrong.

Thanks

 

[Doc Al]

The total force on B is
T - (60 N + w_B) = m_aa

OK, except that it should be m_ba.

and then the total force on B is
T - w_A = -m_aa

I assume you mean this to be the total force on A. You forgot the applied force P (or -P, with your sign convention.)

Since A is accelerating downwards it has a negative acceleration and B has a positive acceleration.

Right: the acceleration of B is +a; of A, -a.

Correct your two equations and combine them to solve for P. (Which I'm guessing is the question.)