When a cube of side length a is in limiting e | vedclass

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

Question

(A) Let the side length of the cube be $a$. When the cube is about to topple,the normal reaction $N$ acts at the edge $A$ of the cube.Taking the torqu

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let the side length of the cube be $a$. When the cube is about to topple,the normal reaction $N$ acts at the edge $A$ of the cube.Taking the torque about the center of mass $G$ of the cube:The torque due to the normal reaction $N$ is $N 	imes (a/2)$.The torque due to the friction force $f = \mu N$ is $f 	imes (a/2) = \mu N 	imes (a/2)$.For the cube to be in rotational equilibrium just before toppling,these torques must balance:$N 	imes (a/2) = \mu N 	imes (a/2)$.Dividing both sides by $N 	imes (a/2)$,we get:$\mu = 1$.

When a cube of side length $a$ is in limiting equilibrium on an inclined plane and it is about to topple,what is the coefficient of friction between the cube and the plane?

Similar Questions

The force required to just move a body up an inclined plane is double the force required to just prevent the body from sliding down the plane. If the coefficient of friction is $\mu$,then the inclination $\theta$ of the plane is:

$A$ body of mass $2 \ kg$ slides down with an acceleration of $4 \ m/s^2$ on an inclined plane having a slope of $30^{\circ}$. The external force required to take the same body up the plane with the same acceleration will be (Acceleration due to gravity $= 10 \ m/s^2$) (in $N$)

$A$ body weighing $20 \ kg$ just slides down a rough inclined plane that rises $5$ in $12$. The coefficient of friction is

$A$ body is projected up along a rough inclined plane of inclination $45^{\circ}$. The coefficient of friction is $0.5$. Then the retardation of the block is

The tension $T$ in the string shown in the figure is ..................... $N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module