Question

A large crate with mass m rests on a horizontal floor. The coefficients of friction between the crate and the floor are \mu _{\rm{s}} and \mu _{\rm{k}}. A woman pushes downward at an angle \theta below the horizontal on the crate with a force \vec F.

A)What magnitude of force \vec F is required to keep the crate moving at constant velocity?

B)If \mu _{\rm{s}} is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. Calculate this critical value of \mu _{\rm{s}}.

Answer 1

In order for the crate to move at constant velocity, all forces must add up to zero.

In the horizontal direction (x), the horizontal component of the push force (Px) must fully oppose the kinetic friction force (F).

In the vertical direction, the normal force (N) will be as large as necessary, to prevent the crate from sinking through the floor. Structural capabilities of floor and crate, of course are required. The normal force must oppose both the weight of the crate (m*g), and the woman's downward component of the push force (Py).

It is of interest for us to develop an equation for the normal force, because the horizontal force of friction is proportional to it.

Component-ize the woman's push force:
Px = P*cos(theta)
Py = P*sin(theta)

Vertical force balance:
N = m*g + Py

Horizontal force balance:
Px = F


Apply trigonometry for Px and Py:
N = m*g + P*sin(theta)
P*cos(theta) = F

Use F = muk*N to find the friction force in terms of normal force:
F = muk*N

Thus:
F = muk*(m*g + P*sin(theta))

Substitute:
P*cos(theta) = muk*(m*g + P*sin(theta))

Multiply through by muk:
P*cos(theta) = muk*m*g + muk*P*sin(theta)

Gather P-terms to one side:
P*cos(theta) - muk*P*sin(theta) = muk*m*g

Factor:
P*(cos(theta) - muk*sin(theta)) = muk*m*g

Solve for P, and our result becomes:
P = muk*m*g/(cos(theta) - muk*sin(theta))
--------------Answer--------------

Notice that for this equation to have physical meaning, the denominator must be a positive number. It cannot be zero, and it cannot be negative.

Solve for where it equals zero:
cos(theta) = muk*sin(theta)
muk = 1/tan(theta)
tan(theta) = 1/muk

It is required that tan(theta) < 1/muk

If tan(theta) were to equal 1/muk, then infinite push force would be required to produce constant speed. In otherwords, the angle would be too steep, that the push force is only increasing the total contact force (friction + normal force as vectors), which is guaranteed to constrain against it.

If tan(theta) were to exceed 1/muk, then the push force would be self-locking, as more push force would be causing it to slow down, rather than to speed up.

Answer 2

guacamole nigger penis