A uniform rod of mass $10 ext{ kg}$ and length $4 ext{ m}$ leans against a smooth perfectly vert

Question

A uniform rod of mass $10 	ext{ kg}$ and length $4 	ext{ m}$ leans against a smooth perfectly vertical wall, making an angle of $45^\circ$ with the rough horizontal floor. The normal reaction force exerted by the wall on the rod is (take $g = 10 	ext{ m/s}^2$)

Professor PC Thomas Classes · Answer

(b) : Let the length of the rod be $L = 4 	ext{ m}$ and its weight be $W = mg = 10 	imes 10 = 100 	ext{ N}$. The rod is in rotational equilibrium. Taking torques about the point of contact with the floor eliminates the torques from static friction and the floor's normal reaction. The torque from the weight acts downwards at distance $(L/2)\cos 45^\circ$. The torque from the wall's normal reaction $N_{wall}$ acts horizontally at height $L\sin 45^\circ$. Equating magnitudes: $N_{wall} 	imes L \sin 45^\circ = W 	imes (L/2) \cos 45^\circ$. Since $\sin 45^\circ = \cos 45^\circ$, this simplifies to $N_{wall} = \frac{W}{2} = \frac{100}{2} = 50 	ext{ N}$.