Question

A massless, frictionless pulley has a rope with two weights suspended from it. The masses of the blocks are as follows: MA = 2m and MB = m. What is the tension in the rope when the blocks are released from rest?

a. 2 mg b. 3/2 mg c. 4/3 mg d. 1/2 mg

Answer 1

a1 a2 Mb Ma

From an above diagram, we have

down positive : m_b g - T = m_b a

T = (m_b g - m_b a)

up positive : T - m_a g = m_a a

(m_b g - m_b a) - m_a g = m_a a

(m_b g - m_a g) = (m_b a + m_a a)

a = (m_b - m_a) g / (m_b + m_a)

using an equation, we have

T = (m_b g - m_b a)

T = m_b g - m_b [(m_b - m_a) g / (m_b + m_a)]

T = m_b g [(m_b + m_a) g / (m_b + m_a)] - m_b g [(m_b - m_a) g / (m_b + m_a)]

T = [m_b (m_b + m_a) - m_b (m_b - m_a) / (m_b + m_a)] g

T = [2 m_b m_a / (m_b + m_a)] g

where, m_b = mass of block B = m

m_a = mass of block A = 2m

then, w get

T = 2 [(m) (2m) / (m) + (2m)] g

T = (4 m² / 3m) g

T = (4/3) mg