Physics, asked by Rishabh496, 1 year ago

In the given arrangement, the normal force applied by block on the ground is​

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Answers

Answered by adrija70
17

Answer:

Fcos(theta)

Explanation:

Normal force applied is the cos component of the force F.

Hope it helps.

Answered by lublana
37

Answer:

mg-F cos\theta

Explanation:

We are given that force applied on the block=F

Mass of block=m

We have to find the normal force applied by block on the ground

When we resolve F then

Vertical component=Fsin\theta

Horizontal component=Fcos\theta

Let N be the normal applied by  the block on the ground

Then , N and Fcos\theta  are applied in the same direction

N+Fcos\theta=mg

N=mg-F cos\theta

Hence, the normal applied on the block=mg-F cos\theta

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