Question 615: Centre of Mass and Rotational Motion

Question

A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of $60^{\circ}$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (Take $g=10 \mathrm{~m} / \mathrm{s}^2$)

Accepted Answer

For translational equilibrium
$$\begin{aligned}
& N_1=M g \
& N_2=f
\end{aligned}$$
For rotational equilibrium
Torque about $A, M g \frac{L}{2} \cos 	heta=N_2 L \sin 	heta$
$$\begin{aligned}
& \frac{M g}{2} \cot 	heta=N_2=f \
& \frac{M g}{2} \cot 30^{\circ}=f \
& \frac{M g}{2} \sqrt{3}=N_2 \
& 100 \sqrt{3}=f
\end{aligned}$$

Answer

200 N

Answer

$200 \sqrt{3} \mathrm{~N}$

Answer

100 N

Answer

$100 \sqrt{3} \mathrm{~N}$

Centre of Mass and Rotational Motion

A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60^{\circ} with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (Take g=10 \mathrm{~m} / \mathrm{s}^2)

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