A mass M text{ kg} is suspended by a weightle | vedclass

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(B) Let $T$ be the tension in the string and $F$ be the horizontal force applied to hold the mass at an angle $	heta = 60^{\circ}$ with the vertical.

Vedclass · Accepted Answer

$M g \sqrt{3}$

Vedclass · Answer

(B) Let $T$ be the tension in the string and $F$ be the horizontal force applied to hold the mass at an angle $	heta = 60^{\circ}$ with the vertical.At equilibrium, the forces acting on the mass are balanced:$1$. Vertical direction: $T \cos 	heta = M g$ (Equation $i$)$2$. Horizontal direction: $T \sin 	heta = F$ (Equation $ii$)Dividing Equation $ii$ by Equation $i$:$\frac{F}{M g} = \frac{T \sin 	heta}{T \cos 	heta} = 	an 	heta$$F = M g 	an 	heta$Given $	heta = 60^{\circ}$, we have:$F = M g 	an 60^{\circ}$Since $	an 60^{\circ} = \sqrt{3}$, the required force is:$F = \sqrt{3} M g$

$A$ mass $M \text{ kg}$ is suspended by a weightless string. The horizontal force required to hold the mass at $60^{\circ}$ with the vertical is

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