Question

A block of mass 2 kg is to be kept at rest against a rough vertical wall (p = 0.4) by applying a horizontal force F. The minimum value of Fis (g = 10 m/s2)​

Answer 1

The minimum value of F = 8 Newtons

Explanation:

Given: A block of mass 2 kg is to be kept at rest against a rough vertical wall (p = 0.4) by applying a horizontal force F.

Find: The minimum value of F

Solution:

acceleration due to gravity g = 10 m/s2

mass m = 2 kg

co-efficient of static friction μ = 0.4

Force acting downwards = mg = 10 * 2 = 20 N

Maximum resisting force = limiting friction

f(lim) = μN = 0.4 * 20 = 8 N

The applied force should be equal to the maximum resisting force to keep the block at rest.

So minimum value of F = 8 N

Answer 2

Given:

A block of mass 2 kg is to be kept at rest against a rough vertical wall (p = 0.4) by applying a horizontal force F.

To find:

Minimum value of F.

Calculation:

When a horizantal force F is being applied to the block , it will be equal to the normal reaction.

So , N = F ......(1)

Now , frictional force will be:

$\therefore \: f = \mu \times N$

$= > f = \mu \times F$

$= > f = \mu F$

Now , this friction must be equal and opposite to the weight in order to be in translational equilibrium:

$\therefore \: f = mg$

$= > \mu F = mg$

$= > (0.4) F =2 \times 10$

$= > (0.4) F =20$

$= > F = \dfrac{20}{0.4}$

$= > F = \dfrac{200}{4}$

$= > F = 50 \: Newton$

So , final answer is:

$\boxed{ \bf{F = 50 \: Newton}}$