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 accleration.

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 accleration.

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.)

 

[quicknote]

for posing your question about masses on a pulley. It seems like you are on the right track with your calculations, but it's possible that there is a mistake in your equations. Here are a few tips to help you solve this problem:

1. Draw a free body diagram for each block, showing all the forces acting on them. This will help you visualize the problem and make sure you have all the forces accounted for.

2. Remember that the tension in a rope is the same throughout the entire length of the rope. So the tension in the rope connecting block B and C is also the same as the tension in the rope connecting A and B.

3. Use Newton's second law (F=ma) to set up equations for each block. Make sure you include the correct signs for the forces (positive for forces in the direction of acceleration, negative for forces in the opposite direction).

4. Once you have set up your equations, solve them simultaneously to find the tension in the rope and the acceleration of the blocks.

I hope this helps you solve the problem and get the correct answer of 190N. If you are still having trouble, try checking your equations and make sure all the forces are accounted for. Good luck!