A block of mass m rests on a horizontal table whose coefficient of static fricti

Question

A block of mass m rests on a horizontal table whose coefficient of static friction is as shown in the figure. Show that, If we want to apply the minimum possible force to move the block, it should be applied with the force pulling upward at angle theta_min = tan^-1 mu_s The minimum force necessary to start the block moving is F_min = mu_s / square root 1 + mu_s^2 mg Once you start the block moving, if you want to apply the least possible force to keep it moving, should you keep the angle at which you are pulling the same, increase it or decrease it?

Explanation / Answer

normal force on block by ground (N) = (mg - Fsin(theta))

frictional force maximum = mu*N

horizontal force = Fcos(theta)

so for to make block move

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

Fsin(theta)+ Fcos(theta)*mu = mu* mg

F = mu*mg/(sin(theta)+ mu*cos(theta))

for F to be minimum

denominator should be maximum

sin(theta) = mu*cos(theta)

theta = tan inverse(mu)

B)

F at that angle = mu*mg /(1+ mu^2)

C)

once we start moving friction coefficient is kinetic friction which is less than static so tan inverse mu term will now be reduced and hence we would wish to decrease the angle

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