Question

4.
A body is kept on inclined plane of inclination30°
then the normal force on body by inclined plane is
(m = 1.732 kg : g = 10 m/s²)
13
1)
2) V3N
N
2
3) 15 N
4) 10/30​

Answer 1

Answer:

(3) 15 N

Explanation:

Given that,

A body of mass 'm' is kept on an inclined plane of inclination 30°.

To find the normal force on the body.

First of all, we need to draw the free body diagram.

$\red{Diagram:-}$ Refer to the attachment.

In the free body Diagram, we can see that, the weight of the body will be Vertically downwards, having magnitude equal to mg where, m is the mass of body and g is acceleratidue to gravity.

Now, breaking this force into it's components, we have,

mg sin 30° will be parallel to the incline and mg cos 30° will be perpendicular to the incline.

Now, the normal force, N which incline is exerting on the body is perpendicular to the incline and opposite to the mg cos30.

Therefore, these two forces will eqaute each other.

Therefore, we will get,

=> N = mg cos 30°

But, it's given that,

• m = 1.732 Kg
• g = 10 m/s^2

And, we know that,

• cos 30 = √3/2 = 1.732/2

Substituting the values, we get,

=> N = 1.732 × 10 × 1.732/2

=> N = (1.732)^2 × 5

=> N = (√3)^2 × 5

=> N = 3 × 5

=> N = 15

Hence, the normal force is (3) 15 N.