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Block $A$ of weight $100 \mathrm{~N}$ rests on a frictionless inclined plane of slope angle $30^{\circ}$. A flexible cord attached to $\mathrm{A}$ passes over a frictionless pulley and is connected to block $B$ of weight $W$. Find the weight $W$ for which the system is in equilibrium.
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Verified Answer
During equilibrium, the force $m g \sin \theta$ acting on block $A$ parallel to the plane should be balanced by the tension in the string, i.e.
$\therefore$ As given that $T=F$,
$m g \sin \theta=T=F$
For block B, $w=T=F$
So, where $w$ is the weight of block $B$.
From above relation then
$$
\begin{aligned}
w &=m g \sin \theta \\
&=100 \times \sin 30^{\circ} \quad(\because m g=100 \mathrm{~N}) \\
&=100 \times \frac{1}{2} \mathrm{~N}=50 \mathrm{~N}
\end{aligned}
$$
For $B$ is at rest, $W=F=50 \mathrm{~N}$ for which the system will be in equilibrium.
$\therefore$ As given that $T=F$,
$m g \sin \theta=T=F$
For block B, $w=T=F$
So, where $w$ is the weight of block $B$.
From above relation then
$$
\begin{aligned}
w &=m g \sin \theta \\
&=100 \times \sin 30^{\circ} \quad(\because m g=100 \mathrm{~N}) \\
&=100 \times \frac{1}{2} \mathrm{~N}=50 \mathrm{~N}
\end{aligned}
$$
For $B$ is at rest, $W=F=50 \mathrm{~N}$ for which the system will be in equilibrium.
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