Question

In the figure shown, all the contact surfaces are smooth and string and pulley are light. When released from rest, it was found that the wedge of mass $m_0$ does not move. Then $\frac{M}{m}$ is

• A. $\sin(37^\circ)$
• B. $\cos(37^\circ)$
• C. $\tan(37^\circ)$
• D. $\cot(37^\circ)$

Answer 1

Answer: (A)

$\sin(37^\circ)$

Answer 2

Since the wedge does not move, the net force on it is zero. For block $m$, the tension $T$ in the string is $T = mg \sin\theta$. For block $M$, the tension $T$ in the string is $T = Mg$. For the wedge to be in equilibrium, the horizontal forces must balance. The horizontal component of the normal force exerted by block $m$ on the wedge is $N_1 \sin\theta = (mg \cos\theta) \sin\theta$. The horizontal component of the tension pulling the pulley is $T \cos\theta$. Thus, $(mg \cos\theta) \sin\theta = T \cos\theta$. This simplifies to $mg \sin\theta = T$. Equating the expressions for $T$, we get $Mg = mg \sin\theta$. Therefore, $\frac{M}{m} = \sin\theta$. Given $\theta = 37^\circ$, $\frac{M}{m} = \sin(37^\circ)$.