How to solve this problem about friction and force applied on an angle?

the coefficient of friction = 0.68
mass = 40 kg
angle at which force F is applied = 60 degrees
take the component of the force in horizontal direction i.e Fcos60
in the horizontal direction two forces are acting viz., static friction and force F. To just start the motion F must be equal to static friction

therefore, Fcos60 =friction
              F x 0.5 = coefficient of friction * normal reaction
              F x 0.5 = 0.68 x 40 x 10
              F = 0.68 x 400 x 2 = 544N

Newton´s 2nd law estates:
Net Force=ma
First you have to solve the situation in the Y axis.
Note in the Y axis there is no displacement, them

Net Force     Normal+Fsin(tetha)-mg=0

or N= mg-Fsin(tetha)
Now the friction force equals =mu*N
Therefore Friction=mu*(mg-Fsin(tetha))

To find the force the worker needs to exert, now you use Newton´s law for the X axis, like this

Fcos(tethe)-friction=0 or friction=Fcos(tetha)

We place zero in the formula, because we just need to know the force necessary to equalise the friction force.

then you have

mu*(mg-Fsin(tetha))=Fcos(tetha)

You finish it. Just solve for F

The second Part of your question, The only difference is that the vertical force exerted by the worker will be oppssite to the Normal force. You tray it step by step