A block of weight W is suspended by a string | vedclass

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(A) Let the tension in the string be $T = 60\,N$. The string makes an angle of $45^{\circ}$ with the vertical.Resolving the tension $T$ into horizonta

Vedclass · Accepted Answer

$\frac{60}{\sqrt{2}}\,N$

Vedclass · Answer

(A) Let the tension in the string be $T = 60\,N$. The string makes an angle of $45^{\circ}$ with the vertical.Resolving the tension $T$ into horizontal and vertical components:Horizontal component $T_x = T \sin 45^{\circ} = 60 	imes \frac{1}{\sqrt{2}} = \frac{60}{\sqrt{2}}\,N$.Vertical component $T_y = T \cos 45^{\circ} = 60 	imes \frac{1}{\sqrt{2}} = \frac{60}{\sqrt{2}}\,N$.For the system to be in equilibrium, the horizontal forces $\overline{F}_1$ and $\overline{F}_2$ must balance the horizontal component of the tension. Assuming $F_1 = F_2 = F$, then $2F = T_x$ is not applicable here as the forces are applied to hold the system. If the forces are applied to balance the horizontal component, $F = T_x = \frac{60}{\sqrt{2}}\,N$.The weight $W$ of the suspended block must balance the vertical component of the tension:$W = T_y = \frac{60}{\sqrt{2}}\,N$.

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