A uniform rod of mass 15,kg and length 5,m is | vedclass

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(C) Let the rod be hinged at the bottom. The length of the rod is $L = 5\,m$ and its mass is $M = 15\,kg$. The weight $Mg = 150\,N$ acts at the center

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$100$

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(C) Let the rod be hinged at the bottom. The length of the rod is $L = 5\,m$ and its mass is $M = 15\,kg$. The weight $Mg = 150\,N$ acts at the center of mass,which is at a distance of $L/2 = 2.5\,m$ from the hinge.From the figure,the height of the string attachment is $3\,m$. Since the rod length is $5\,m$,the angle $	heta$ that the rod makes with the vertical wall can be found. However,looking at the geometry,the string is horizontal. Let $\alpha$ be the angle the rod makes with the vertical wall. Then $\cos \alpha = 3/5 = 0.6$ and $\sin \alpha = 4/5 = 0.8$.Taking torque about the hinge point:Clockwise torque due to weight = $Mg 	imes (L/2) \sin \alpha = 150 	imes 2.5 	imes 0.8 = 300\,N\cdot m$.Counter-clockwise torque due to tension $T$ = $T 	imes L \cos \alpha = T 	imes 5 	imes 0.6 = 3T$.Equating the torques for rotational equilibrium:$3T = 300$$T = 100\,N$.

$A$ uniform rod of mass $15\,kg$ and length $5\,m$ is held stationary with the help of a light string as shown. The tension in the string is ......... $N.$

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