Question

Two smooth blocks of masses $m$ and $m'$ connected by a light inextensible strings are moving on a smooth wedge of mass $M$. If a force $F$ acts on the wedge the blocks do not slide relative to the wedge. Find the (a) acceleration of the wedge and (b) value of $F$.

Answer 1

(a) Acceleration of the wedge: $a = \frac{m'g}{m}$

(b) Value of $F$: $F = \frac{m'g}{m}(M + m + m')$

Answer 2

The problem asks us to find the acceleration of the wedge and the applied force $F$ under the condition that the blocks do not slide relative to the wedge.

Understanding the Constraint:

The crucial condition "the blocks do not slide relative to the wedge" implies that all three bodies (block $m$, block $m'$, and wedge $M$) move together as a single system in the horizontal direction. Let their common horizontal acceleration be $a$.

Part (a): Acceleration of the wedge

1. Free Body Diagram for block $m$:

   Block $m$ is on the horizontal surface of the wedge. The only horizontal force acting on it is the tension $T$ from the string. Applying Newton's second law in the horizontal direction for block $m$: $T = m a \quad \text{(Equation 1)}$

2. Free Body Diagram for block $m'$:

   Block $m'$ is hanging vertically. The forces acting on it are the tension $T$ in the string (upwards) and its weight $m'g$ (downwards). Since block $m'$ does not slide relative to the wedge, its vertical acceleration must be zero (as the wedge itself has no vertical acceleration). Applying Newton's second law in the vertical direction for block $m'$: $T - m'g = 0$ $T = m'g \quad \text{(Equation 2)}$

3. Solving for acceleration $a$:

   Equating the expressions for tension $T$ from Equation 1 and Equation 2: $ma = m'g$ Therefore, the acceleration of the wedge (and the blocks) is: $a = \frac{m'g}{m}$

Part (b): Value of Force $F$

1. Consider the entire system:

   The external horizontal force acting on the combined system (wedge $M$ + block $m$ + block $m'$) is $F$. The total mass of this combined system is $(M + m + m')$. Applying Newton's second law to the entire system in the horizontal direction: $F = (M + m + m') a$

2. Substitute the value of $a$:

   Substitute the expression for $a$ found in Part (a) into this equation: $F = (M + m + m') \left(\frac{m'g}{m}\right)$ $F = \frac{m'g}{m}(M + m + m')$

The condition that blocks do not slide relative to the wedge implies a common horizontal acceleration $a$ for all three masses. By analyzing the forces on block $m$ horizontally ($T=ma$) and on block $m'$ vertically ($T=m'g$), we find $a = m'g/m$. Then, considering the entire system, the net external force $F$ causes this common acceleration, so $F = (M+m+m')a$. Substituting $a$ gives the value of $F$.