Question

A block of mass is pressed into a vertical wall with force 'F' as shown in the figure. If the block stays at rest friction between block and wall can be (A) mg - F cose upward (B) Fcos 0 - mg downward (C) Zero (D) mg + F cos e​

Answer 1

First of all, see the Free Body Diagram as shown in attached image.

• Force F can be divided into perpendicular components as shown in the 2nd diagram.

Since the block is in equilibrium (i.e. at rest), we can say:

$\rm \therefore F \sin( \theta) + f = mg$

• 'f' is frictional force directed upwards.

$\rm \implies f = mg - F \sin( \theta)$

So, final answer is :

$\boxed{ \bf f = mg - F \sin( \theta) }$

Answer 2

Explanation:

First of all, see the Free Body Diagram as shown in attached image.

Force F can be divided into perpendicular components as shown in the 2nd diagram.

Since the block is in equilibrium (i.e. at rest), we can say:

\rm \therefore F \sin( \theta) + f = mg∴Fsin(θ)+f=mg

'f' is frictional force directed upwards.

\rm \implies f = mg - F \sin( \theta) ⟹f=mg−Fsin(θ)

So, final answer is :

\boxed{ \bf f = mg - F \sin( \theta) }

f=mg−Fsin(θ)