The minimum possible force to move the block has to be just equal to the maximum possible value of static friction. (Static friction is the friction force between the table and the block that keeps the block from moving.) The maximum possible value of the magnitude of the static friction is
`F_s = mu_sN` , where N is the normal force (the force from the table on the block, acting perpendicular to the table.)
If the horizontal component of the applied force is just equal to the maximum possible static friction, the block will start moving with no acceleration, so from the second Newton's Law,
`Fcos(theta) = mu_sN`
The normal force can also be found from the second Newton's Law, considering the components of the forces perpendicular to the table:
`-W + Fsin(theta) + N = 0`
Here, W is the magnitude of the gravitational force on the block, or weight.
From here, `N = W - Fsin(theta) `
and `Fcos(theta) = mu_s(W - Fsin(theta))`
Solving this for F results in
`F = (mu_sW)/(cos(theta) + mu_ssin(theta))`
To find the minimum value of F, take the derivative with respect to the angle:
`(dF)/(d(theta)) = mu_sW*(-sin(theta)+mu_scos(theta))/(cos(theta)+mu_ssin(theta))^2`
The derivative is zero when ` `
`mu_s = tan(theta)` , which means that the minimal value of the force is reached when
`theta = arctan(mu_s) = arctan(0.6) = 31 degrees `
The angle of approximately 31 degrees is required to move the block with the minimal force possible.
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`
`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:
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
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` .
This equation is satisfied when the numerator is equal to zero. Therefore:
Solve for `theta` .
`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:
`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.