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

$A$ charged particle moves through a magnetic field perpendicular to its direction. Then

For a reaction,$A + B^{2+} \rightarrow B + A^{2+}$; $E^{\circ} = 0.2955 \ V$. Hence,the equilibrium constant of the reaction at $25^{\circ}C$ is:

An infinitely long thin straight wire has a uniform linear charge density of $\frac{1}{3} ~Cm^{-1}$. The magnitude of the electric field intensity at a point $18 ~cm$ away is (given $\varepsilon_0 = 8.85 \times 10^{-12} ~C^2 N^{-1} m^{-2}$):

The equation of a plane containing the line of intersection of the planes $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and passing through the point $(1, 1, 0)$ is:

Which of the following is an azo dye?

Vedclass Test Series

Exam Paper Generator

Online Exam Module