Browse Questions

# Consider the following Atwood machine with three masses in the ratio m$_1$: m$_2$: m$_3$:: 1:2:3, hung with a massless string over a friction-less pulley. What is the tension in the string between masses m$_1$ and m$_2$?

Answer: $\large\frac{5}{3}$$mg The Free Body diagrams for the problem can be drawn as follows: Net force in the direction of motion of m_1 is F_1 = T- m_1g =m_1a as per Newton's second law. Net force in the direction of motion of m_2 is F_1 = T- T'=m_2a as per Newton's second law. Net force in the direction of motion of m_3 is F_3 = m_3g- T'=m_3a as per Newton's second law. Solving for acceleration, we get: a = \large\frac{m_2 + m_3 - m_1}{m_1+m_2+m_3}$$g = \large\frac{2m+3m-m}{m+2m+3m}$$g = \large\frac{2}{3}$$g$
Now, $T-m_1g' = m_1 a \rightarrow T = m_1 (g+a)$.
Subsituting for $a$ and for $m_1 = m$, we get: $T = m (g + \large\frac{2}{3}$$g) = \large\frac{5}{3}$$mg$