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(A) Let $T$ be the tension in the string $AB$. At point $B$,the forces acting are: $1$. Tension $T$ in string $AB$ acting at an angle $	heta$ with th

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$F / \sin 	heta$

Vedclass · Answer

(A) Let $T$ be the tension in the string $AB$. At point $B$,the forces acting are: $1$. Tension $T$ in string $AB$ acting at an angle $	heta$ with the vertical. $2$. Horizontal force $F$ acting towards the right. $3$. Tension $T'$ in the vertical string supporting mass $M$,where $T' = Mg$. Resolving the tension $T$ into horizontal and vertical components: Horizontal component: $T \sin 	heta$ Vertical component: $T \cos 	heta$ For equilibrium at point $B$: Horizontal forces: $T \sin 	heta = F$ Vertical forces: $T \cos 	heta = Mg$ From the horizontal equilibrium equation,we get: $T = \frac{F}{\sin 	heta}$

$A$ mass $M$ is suspended by a rope from a rigid support at $A$ as shown in the figure. Another rope is tied at point $B$,and it is pulled horizontally with a force $F$. If the string $AB$ makes an angle $\theta$ with the vertical in equilibrium,then the tension in the string $AB$ is:

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