Question

Figure shows three blocks attached by cords that

loop over frictionless pulleys. Block B lies on a
frictionless table; the masses are ma = 6.00 kg.
mg = 8.00 kg, and mc = 10.0 kg. When the blocks are
released, what is the tension in the cord at the right?

Answer 1

Answer: The tension in the cord at the right is 81.7N.

Step-by-step explanation:

In order to solve this we consider a system with a clockwise motion considered as positive which means that for a downward position for block C is positive and for block B rightward is positive and for Block A upward is positive)

Formula = mC g- mA g = M a (where M = mass of system = 24.0kg)

a(acceleration) = g(mC-mA)/M = 1.63m/s^2

Force on Block C = mCg -T = mCa

T = mCg(2mA+mB)/M

T = 81.7N

Hence Tension = 81.7N