Solution to Question 23
Step 1: Analyze Directions and Forces
Since $m_2 > m_1$, the normal force from $m_2$ will dominate, pushing the wedge to the Left. Let the acceleration of the wedge be $a$ (towards the left).
In the non-inertial frame of the wedge (accelerating Left), a pseudo force $ma$ acts to the Right on both blocks.
Step 2: Normal Forces on the Wedge
The wedge moves horizontally due to the horizontal components of the normal forces exerted by the blocks.
- From Right Block ($m_2$): The normal force $N_2$ acts perpendicular to the surface (down-left). Its horizontal component pushes the wedge Left.
- From Left Block ($m_1$): The normal force $N_1$ acts perpendicular to the surface (down-right). Its horizontal component pushes the wedge Right.
Equation of Motion for the Wedge (Net Force = Mass × Acceleration):
$$(N_2 – N_1) \sin \theta = m_0 a \quad \dots (1)$$Step 3: Calculate Normal Forces ($N_1, N_2$)
We analyze forces perpendicular to the incline for each block, including the Pseudo Force ($ma$, acting Right).
For Block $m_1$ (Left Slope):
The pseudo force acts to the Right, pushing the block into the incline.
For Block $m_2$ (Right Slope):
The pseudo force acts to the Right, pulling the block away from the incline.
Step 4: Solve for Acceleration
Substitute expressions (2) and (3) into equation (1):
$$[(m_2 g \cos \theta – m_2 a \sin \theta) – (m_1 g \cos \theta + m_1 a \sin \theta)] \sin \theta = m_0 a$$Group the gravity terms and acceleration terms:
$$(m_2 – m_1) g \sin \theta \cos \theta – (m_2 + m_1) a \sin^2 \theta = m_0 a$$Move all terms with $a$ to the right side:
$$(m_2 – m_1) g \sin \theta \cos \theta = m_0 a + (m_1 + m_2) a \sin^2 \theta$$ $$(m_2 – m_1) g \sin \theta \cos \theta = a [m_0 + (m_1 + m_2) \sin^2 \theta]$$