A block placed on a rough inclined plane of i | vedclass

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

Question

(C) In the first case,the block is on the verge of moving upwards,so the friction force acts downwards. The force balance equation is: $F = mg \sin 30

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) In the first case,the block is on the verge of moving upwards,so the friction force acts downwards. The force balance equation is: $F = mg \sin 30^{\circ} + \mu mg \cos 30^{\circ} \dots (1)$In the second case,the block is on the verge of sliding downwards,so the friction force acts upwards. The force balance equation is: $F + \mu mg \cos 60^{\circ} = mg \sin 60^{\circ} \implies F = mg \sin 60^{\circ} - \mu mg \cos 60^{\circ} \dots (2)$Since the magnitude of $F$ is the same in both cases,equate $(1)$ and $(2)$:$mg \sin 30^{\circ} + \mu mg \cos 30^{\circ} = mg \sin 60^{\circ} - \mu mg \cos 60^{\circ}$$\mu (\cos 30^{\circ} + \cos 60^{\circ}) = \sin 60^{\circ} - \sin 30^{\circ}$$\mu (\frac{\sqrt{3}}{2} + \frac{1}{2}) = \frac{\sqrt{3}}{2} - \frac{1}{2}$$\mu = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Similar Questions

$A$ truck starting from rest moves with an acceleration of $5 \, m/s^2$ for $1 \, s$ and then moves with constant velocity. The velocity $w.r.t$ ground $v/s$ time graph for the block in the truck is (Assume that the block does not fall off the truck and the coefficient of friction $\mu = 0.2$):

$A$ small block of mass $0.1 \ kg$ lies on a fixed inclined plane $PQ$ which makes an angle $\theta$ with the horizontal. $A$ horizontal force of $1 \ N$ acts on the block through its center of mass as shown in the figure. The block remains stationary if (take $g = 10 \ m/s^2$):

In the figure,a ball of mass $m$ is tied with two strings of equal length as shown. If the rod is rotated with angular velocity $\omega$,then

$STATEMENT-1$: It is easier to pull a heavy object than to push it on a level ground.
$STATEMENT-2$: The magnitude of frictional force depends on the nature of the two surfaces in contact.

Vedclass Test Series

Exam Paper Generator

Online Exam Module