Question

A mass $m$ is suspended by a rope from a rigid support at P as shown in the figure. Another rope is tied at the end Q which is pulled horizontally with a force $F$. If the rope PQ makes angle $\theta$ with the vertical, then tension in the string PQ is

• A. $F\sin \theta$
• B. $\frac{F}{\sin \theta }$
• C. $F\cos \theta$
• D. $\frac{F}{\cos \theta }$

Answer 1

The correct option is B $\frac{F}{\sin \theta }$

Let us consider all the forces acting at the point Q.

Taking components of $T$ in vertical and horizontal direction, we get;
$T_x=T\sin \theta \text{and}{T}_y=T\cos \theta$
For horizontal equilibrium:
$T\sin \theta =F\Rightarrow T=\frac{F}{\sin \theta }$