Physics - NLM Question with Solution | TestHub

Consider the equilibrium of point Q. The forces acting at point Q are:
1. Tension $T$ in the string PQ, acting at an angle $\theta$ with the vertical.
2. Horizontal force $F$ pulling to the right.
3. Downward force due to the suspended mass $M$, which is $Mg$.

Resolving the tension $T$ into horizontal and vertical components:
Horizontal component of tension = $T\sin \theta$ (acting towards the left)
Vertical component of tension = $T\cos \theta$ (acting upwards)

For the system to be in equilibrium, the net force in the horizontal direction must be zero. Therefore, the horizontal component of the tension must balance the applied horizontal force $F$:
$T\sin \theta =F$

Rearranging the equation to solve for $T$, we get:
$T=\frac{F}{\sin \theta }$

Answer: $\frac{F}{\sin \theta }$