An equilateral prism of mass m rests on a rou | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To topple the prism, it must rotate about the edge $C$.The force $F$ is applied at the top vertex $A$ at a height $h = a\frac{\sqrt{3}}{2}$ from t

Vedclass · Accepted Answer

$\frac{mg}{\sqrt{3}}$

Vedclass · Answer

(A) To topple the prism, it must rotate about the edge $C$.The force $F$ is applied at the top vertex $A$ at a height $h = a\frac{\sqrt{3}}{2}$ from the base.The weight $mg$ acts at the center of mass, which is at a horizontal distance of $a/2$ from the edge $C$.For the prism to be on the verge of toppling, the torque due to force $F$ about point $C$ must balance the torque due to gravity $mg$ about point $C$.Torque due to $F$ about $C$ = $F 	imes h = F 	imes a\frac{\sqrt{3}}{2}$.Torque due to $mg$ about $C$ = $mg 	imes \frac{a}{2}$.Equating the torques: $F 	imes a\frac{\sqrt{3}}{2} = mg 	imes \frac{a}{2}$.Solving for $F$, we get $F = \frac{mg}{\sqrt{3}}$.

Similar Questions

There is a rod of length $l$. The velocities of its two ends are $v_1$ and $v_2$ in opposite directions normal to the rod. The distance of the instantaneous axis of rotation from $v_1$ is:

$A$ hollow cone of radius $R$ and height $H = 2R$ is placed on an inclined plane of inclination $\theta$. If $\theta$ is increased gradually,at what value of $\theta$ will the cone topple? Assume sufficient friction is present to prevent slipping.

The instantaneous velocity of point $B$ of a rod of length $0.5 \ m$ is $3 \ m/s$ at an angle of $30^{\circ}$ with the rod as shown. Find the angular velocity of the rod such that the velocity of end $A$ is minimum.

$A$ block with a square base measuring $a \times a$ and height $h$ is placed on an inclined plane. The coefficient of friction is $\mu$. The angle of inclination $(\theta)$ of the plane is gradually increased. The block will:

Vedclass Test Series

Exam Paper Generator

Online Exam Module