Question

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

Answer 1

Answer:

Fcos(theta)

Explanation:

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

Hope it helps.

Answer 2

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=$mg-F cos\theta$