In the figure,the tension in the diagonal str | vedclass

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(A) Let the tension in the diagonal string be $T = 60\,N$. The string is at an angle of $45^{\circ}$ with the horizontal.Resolving the tension $T$ int

Vedclass · Accepted Answer

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

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(A) Let the tension in the diagonal string be $T = 60\,N$. The string is at an angle of $45^{\circ}$ with the horizontal.Resolving the tension $T$ into horizontal and vertical components:Horizontal component $T_x = T \cos 45^{\circ} = 60 	imes \frac{1}{\sqrt{2}} = \frac{60}{\sqrt{2}}\,N$.Vertical component $T_y = T \sin 45^{\circ} = 60 	imes \frac{1}{\sqrt{2}} = \frac{60}{\sqrt{2}}\,N$.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:$F_1 = F_2 = T_x = \frac{60}{\sqrt{2}}\,N$.The weight $W$ of the suspended block must balance the vertical component of the tension:$W = T_y = \frac{60}{\sqrt{2}}\,N$.

In the figure,the tension in the diagonal string is $60\,N$. Find the magnitude of the horizontal forces $\overline{F}_1$ and $\overline{F}_2$ that must be applied to hold the system in the position shown. What is the weight of the suspended block?

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