Physics, asked by Ashishmishra5821, 9 months ago

If the block a does not move with respect to the wedge the normal reaction will be

Answers

Answered by Afreenakbar

Answer:

The normal force exerted by the wedge on block A is given by

$N = \frac{mg}{(cos \theta + \mu sin \theta )}$

Explanation:

The net force acting on block A must be zero if it is at rest in relation to the wedge. Both block A's weight (in milligrams) and the wedge's normal force (in Newtons) are operating on it.

We may construct the equations of motion for the x and y directions using Newton's second law (F=ma):

In the x direction:

Nsinθ - μNcosθ = 0

In the y direction:

Ncosθ + mg - R = 0

where:

θ is the angle of the wedge
μ is the coefficient of static friction between the block and the wedge
R is the reaction force exerted by the ground on the wedge

Since the block A is not able to moving with respect to the wedge, the net force in both x and y directions is to be zero.

Therefore, we can solve the above equations for the normal force N and the reaction force R:

$N = \frac{mg}{(cos \theta + \mu sin \theta )}$

R = mg tanθ

So, the normal force exerted by the wedge on block A is given by

$N = \frac{mg}{(cos \theta + \mu sin \theta )}$ .