Question

_{​Three blocks are attached by a cord that loop over frictionless pulleys. Block B lies on a table with u=0.1; the masses are mA= 5kg, mB=10kg, and mc=20kg. When the blocks are released, what is the tension in the cord at the right?}

_{​Please outline your work in details.}

Answer 1

the net force moving the system to the side of C is

Fnet = (20.0 kg - 5.0 kg) *9.81 m/s^2 = 147.15 N

The system's total inertial mass is 35 kg. Hence the system as a whole will accelerate to the rside of C with:

a = F/m = 147.15 N /35 kg = 4.20 m/s^2

From this we can obtain all the tensions:

left side string:

T1 - 5.0 kg * 9.81 m/s^2= 5.0kg * 4.20 m/s^2 => T1 = 70.05 N

right side string:

T2 - T1 = 10.0 kg * 4.20 m/s^2 => T2 = 112.05 N

I hope help you.