Question

A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q inclined at an angle. to the vertical. The block will remain in equilibrium, if the coefficient of friction between it and the surface

Answer 1

see figure,
here we have to break force Q into two components .
e.g., $\bf{Q_{horizontal}=Qsin\theta}$
and $\bf{Q_{vertical}=Qcos\theta}$
of course we can write in vector format,
e.g., $\bf{\vec{Q}=Qsin\theta\hat{i}-Qcos\theta\hat{j}}$

we know, friction is a contact force and it is directly proportional to normal reaction between between two contacting bodies.
here, force P and horizontal component of Q act forward then, friction acts backward direction.

hence, friction = P + Qsinθ
fr = μN = P + Qsinθ

here normal reaction , N = Mg + Qcosθ
so, μ = (P + Qsinθ)/N
= (P + Qsinθ)/(Mg + Qcosθ)

Answer 2

=》Here we have to break force Q into two components .

e.g., Q horizontal = Q sin θ

and Q vertical = Q cos θ

we know, friction is a contact force and it is directly proportional to normal reaction between between two contacting bodies.

here, force P and horizontal component of Q act forward then, friction acts backward direction.

hence, friction = P + Qsinθ

fs = μN = P + Qsinθ

here normal reaction , N = Mg + Qcosθ

so, μ = (P + Qsinθ)/N

= (P + Qsinθ)/(Mg + Qcosθ)

》》Answer《《