The figure below shows three blocks attached by cords that loop over frictionles
ID: 2133298 • Letter: T
Question
The figure below shows three blocks attached by cords that loop over frictionless pulleys. Block B lies on a frictionless table; the masses are mA = 3.80 kg, mB = 8.00 kg, mC = 13.0 kg. When the blocks are released, what is the tension in the cord at the right?
N
Explanation / Answer
let TL is the tension in the left cord and TR is the
tension in the right cord
net force on A,
Fa = TL - ma*g
ma*a = TL - ma*g --(1)
net force on c,
Fc = mc*g-TR
mc*a = mc*g - TR --(2)
net force on B(in forizonatl direction),
Fb = TR - TL
mb*a = TR - TL --(3)
add eqns 1,2 and 3 we get
a*(ma+mb+mc) = mc*g - ma*g
a = (mc*g - ma*g)/(ma+mb+mc) = 3.65 m/s^2
from eqn (1) TL = ma*a + ma*g = 51.05 N
from eqn (2) TR = mc*g - mc*a = 79.95 N