## A block of mass m is at rest under the action of force F against a wall as shown in figure…

Q: A block of mass m is at rest under the action of force F against a wall as shown in figure. Which of the following statement is incorrect? (a)f=mg (where f is the frictional force)

(b)F=N (where N is the normal force)

(c)F will not produce torque

(d)N will not produce torque

Ans: (d)

Sol: For equilibrium ,

F = N

f = mg

and $\displaystyle \tau_N + \tau_f = 0$

Hence , τf ≠ 0

τN ≠ 0

## What is the maximum value of the force F such that the block shows in the arrangement, does not move?

Q: What is the maximum value of the force F such that the block shows in the arrangement, does not move? (a)20 N

(b)10 N

(c)12 N

(d)15 N

Ans: (a)

Sol: For vertical equilibrium of block

$\large N = mg + F sin60$

$\large N = \sqrt{3}g + \frac{\sqrt{3}F}{2}$

For No motion

$\large f \ge F cos60$

$\large \mu N \ge F cos60$

$\large \frac{1}{2\sqrt{3}}(\sqrt{3}g + \frac{\sqrt{3}F}{2}) \ge \frac{F}{2}$

$\large g \ge F/2$

$\large F \le 2 g$

= 20 N

## An insect crawl up a hemispherical surface very slowly (see the figure). The coefficient of friction between…

Q: An insect crawl up a hemispherical surface very slowly (see the figure). The coefficient of friction between the surface and the insect is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle α with the vertical, the maximum possible value of α is given (a) cot α =3

(b) tan α =3

(c) sec α =3

(d) cosec α =3

Ans: (a)

Sol:  N = m g cosα …(i)

f = m g sinα  …(ii)

$\large \frac{f}{N} = tan\alpha$

$\large \mu= tan\alpha$

$\large \frac{1}{3} = tan\alpha$

## A long horizontal rod has a bead which can slide along its length and is initially placed at a distance L from one end A of the rod…

Q: A long horizontal rod has a bead which can slide along its length and is initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with a constant angular acceleration α , if the coefficient if friction between the rod and bead is μ, and gravity is neglected, then the time after which the bead starts slipping is

(a)√(μ/α)

(b)μ/√α

(c)1/√μα

(d)Infinitesimal

Ans: (a)

## A block of mass 0.1 kg is held against a wall applying a horizontal force of 5 N on the block…

Q: A block of mass 0.1 kg is held against a wall applying a horizontal force of 5 N on the block. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is

(a)2.5 N

(b)0.98 N

(c)4.9 N

(d)0.49 N

Ans: (b)

Sol: fmax = u N = 0.5 × 5= 2.5 N

Force due to weight = mg = 0.1×9.8 = 0.98 N

As , 0.98 < 2.5 Hence , frictional force = 0.98 N