Question

A block of mass m1 is suspended vertically with a rope connected to a block of mass m2. The block of mass m2 is on a frictionless plane inclined at theta with respect to the horizontal. If m1=2kg, the acceleration of the block is 1.96 m/s^2 downward. If m1=1kg the acceleration of the block is 1.22 m/s^2 upwards. Determine theta (angle) and m2.

Answer 1

$\begin{cases} m_1\vec a=m_1\vec g+\vec T, \\ m_2\vec a=m_2\vec g+\vec T; \end{cases}$

$m_2(g\sin \theta±a)=m_1(g\mp a),$

$\begin{cases} m_2(9.81\sin\theta+1.96)=2\cdot(9.81-1.96), \\ m_2(9.81\sin\theta-1.22)=1\cdot(9.81+1.22); \end{cases}$

where

$\theta=\arcsin0.89=62.87°,$

$m_2=1.47~kg.$