The coefficient of static friction, mu _s bet | vedclass

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

Question

Let the mass of the block $B$ is $M$.

In equilibrium of block $B$ implies $T=M g \ldots(i)$

If block $A$ does not move, then $T=f_{s}$

where

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

Let the mass of the block $B$ is $M$.

In equilibrium of block $B$ implies $T=M g \ldots(i)$

If block $A$ does not move, then $T=f_{s}$

where $f_{s}=$ frictional force $=\mu_{S} N=\mu_{s} m g$

implies $T=\mu_{s} m g \ldots(i i)$

Thus, from Eqs. $( i )$ and $(ii),$ we have

$M=\mu_{s} m g$ or $M=\mu_{s} m$

Given: $\mu_{s}=0.2, m=2 k g$

$	herefore M=0.2 	imes 2=0.4 k g$

Similar Questions

The limiting value of static friction between two contact surfaces is ...........

A block of mass $1\, kg$ is at rest on a horizontal table. The coefficient of static friction between the block and the table is $0.5.$ The magnitude of the force acting upwards at an angle of $60^o$ from the horizontal that will just start the block moving is

A body takes $1\frac{1}{3}$ times as much time to slide down a rough identical but smooth inclined plane. If the angle of inclined plane is $45^o$, the coefficient of friction is

A block of mass $m$ is placed on a surface having vertical cross section given by $y=x^2 / 4$. If coefficient of friction is $0.5$ , the maximum height above the ground at which block can be placed without slipping is:

Someone is using a scissors to cut a wire of circular cross section and negligible weight. The wire slides in the direction away from the hinge until the angle between the scissors blades becomes $2 \alpha$. The friction coefficient between the blades and the wire, is :-