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.
http://hyperphysics.phy-astr.gsu.edu/hbase/N2st.html
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.
https://www.physicstutorials.org/home/mechanics/dynamics/newtons-second-law-of-motion
https://www.themathpage.com/
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