A rectangular block has a square base measuri | vedclass

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

Question

(D) Let $C$ be the center of mass of the block. The height of the center of mass from the base is $h/2$.For the block to topple,the torque due to the

Vedclass · Accepted Answer

$\mu > \frac{a}{2h}$

Vedclass · Answer

(D) Let $C$ be the center of mass of the block. The height of the center of mass from the base is $h/2$.For the block to topple,the torque due to the frictional force $f$ about the edge of the base must be greater than the torque due to the normal force $N$ about the same edge.Taking torque about the front edge of the base:$	au_{f} = f 	imes h$$	au_{N} = N 	imes \frac{a}{2}$For toppling,$	au_{f} > 	au_{N} \implies f 	imes h > N 	imes \frac{a}{2}$.Since the block is moving,$f = \mu N$.Substituting this,$\mu N h > N \frac{a}{2}$.$\mu h > \frac{a}{2} \implies \mu > \frac{a}{2h}$.Thus,the block will topple if $\mu > \frac{a}{2h}$.

$A$ rectangular block has a square base measuring $a \times a$ and its height is $h$. It moves on a horizontal surface in a direction perpendicular to one of the edges. The coefficient of friction is $\mu$. It will topple if

Similar Questions

$A$ $2 \, kg$ block on ice is given a velocity of $6 \, m/s$. If it comes to rest in $10 \, s$,what is the coefficient of friction?

In the given figure,the acceleration of the block of mass $M = \frac{10}{3} \, kg$ is (given $g = 10 \, ms^{-2}$,$\mu = \frac{1}{3}$,$F = 50 \, N$,and $\theta = \sin^{-1}(\frac{3}{5})$):

The coefficient of friction between the block and the surface is $0.03$. Find the acceleration of the system in $m/s^{2}$. $(g = 10\,m/s^{2})$

Consider a car moving on a straight road with a speed of $100\, m/s$. The distance at which the car can be stopped is ........ $m$. $[\mu_k = 0.5, g = 10\, m/s^2]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module