The force required to move a body up a rough | vedclass

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

Question

(C) For upward motion,the force required is $F_{up} = mg(\sin 	heta + \mu \cos 	heta)$.For downward motion,the force required to prevent sliding is

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}}$

Vedclass · Answer

(C) For upward motion,the force required is $F_{up} = mg(\sin 	heta + \mu \cos 	heta)$.For downward motion,the force required to prevent sliding is $F_{down} = mg(\sin 	heta - \mu \cos 	heta)$.According to the problem,$F_{up} = 2 F_{down}$.Substituting the expressions,we get:$mg(\sin 	heta + \mu \cos 	heta) = 2 mg(\sin 	heta - \mu \cos 	heta)$$\sin 	heta + \mu \cos 	heta = 2 \sin 	heta - 2 \mu \cos 	heta$$3 \mu \cos 	heta = \sin 	heta$$\mu = \frac{1}{3} 	an 	heta$Given $	heta = 60^{\circ}$,we have $	an 60^{\circ} = \sqrt{3}$.Therefore,$\mu = \frac{1}{3} 	imes \sqrt{3} = \frac{1}{\sqrt{3}}$.

The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction,when the angle of inclination of the plane is $60^{\circ}$,is

Similar Questions

If $(1, a)$ and $(b, 2)$ are conjugate points with respect to the circle $x^2+y^2=25$,then $4a+2b$ is equal to

If the circle $x^2+y^2+8x-4y+c=0$ touches the circle $x^2+y^2+2x+4y-11=0$ externally and cuts the circle $x^2+y^2-6x+8y+k=0$ orthogonally,then $k$ is equal to

Which of the following is a square planar complex ion?

In male cockroaches,sperms are stored in which part of the reproductive system?

Magnetic induction at the centre of a circular loop of area $\pi \ m^2$ is $0.1 \ T$. The magnetic moment of the loop is ( $\mu_0 = \text{permeability of air}$ ).

Vedclass Test Series

Exam Paper Generator

Online Exam Module