The static force opposes the motion of the object, and the maximum value of the static-friction force is proportional to the normal force `F_n` . The normal force is equal to the vertical component of the force `F` . Keeping the magnitude of `F` constant and increasing theta from zero results in an increase in the vertical component of the force and a decrease in `F_n` ; thus decreasing the maximum static-friction force `f_(max)` . The object will begin to move if the horizontal component of the force exceeds `f_(max)` .

Apply Newton's second law to the block and solve for `F` in terms of `theta` .

`eq. (1) :->` `sum F_x=Fcos(theta)-f_s=0`

and

`eq. (2) :->` `sum F_y=F_n+Fsin(theta)-mg=0`

The maximum magnitude of force that can be applied before the block slips is:

`f_s=f_(s,max)=mu_s*F_n`

Now eliminate `f_s` and `F_n` in `eq. (1)` and `eq. (2)` and solve for `F` .

You will find that the force as a function of angle is

`F(theta)=(mu_smg)/(cos(theta)+mu_ssin(theta))`

Now to find the minimum force we must take the derivative and set it equal to zero to find the critical value of theta between `0` and `pi/2` .

`(dF(theta))/(d theta)=-(mu_smg)(-sin(theta)+mu_scos(theta))/(cos(theta)+mu_ssin(theta))^2=0`

This equation is satisfied when the numerator is equal to zero. Therefore:

`(dF(theta))/(d theta)=-(mu_smg)(-sin(theta)+mu_scos(theta))=0`

`(-sin(theta)+mu_scos(theta))=0`

Solve for `theta` .

`sin(theta)=mu_scos(theta)`

`tan(theta)=mu_s`

`theta=tan^-1(mu_s)+pi*n` , `n` = any integer

Although we only care about solutions between `0` and `pi/2` so let `n=0` . Then we have one solution which is:

`tan^-1(0.6)~~pi/6`

`F(pi/6)=(mu_smg)/(cos(pi/6)+mu_ssin(pi/6))`

`F(pi/6)=(240 N)/(sqrt(3)/2+0.6*1/2)~~206 N`

Hence at an angle of about `pi/6` it only takes about `206 N` to move the box, which is the minimum possible value.