A mass 'M' is suspended by a rope from a rigi | vedclass

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(B) At point '$Q$',the forces acting are: $1$. Tension '$T$' in string '$PQ$' acting at an angle '$	heta$' with the vertical. $2$. Horizontal force '

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$\frac{F}{\sin 	heta}$

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(B) At point '$Q$',the forces acting are: $1$. Tension '$T$' in string '$PQ$' acting at an angle '$	heta$' with the vertical. $2$. Horizontal force '$F$' acting to the right. $3$. Tension '$T_2$' in the vertical string supporting mass '$M$',which is equal to '$Mg$'. For the system to be in equilibrium,the sum of horizontal and vertical forces must be zero. Resolving '$T$' into components: Horizontal component: $T \sin 	heta = F$ Vertical component: $T \cos 	heta = Mg$ From the horizontal component equation,we get: $T = \frac{F}{\sin 	heta}$ Therefore,the tension in the string '$PQ$' is $\frac{F}{\sin 	heta}$.

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

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