Question

A large crate with mass m rests on a horizontal
floor. The coefficients of friction between the
crate and the floor are us and uie. A woman
pushes downwards at an angle = 60° below
the horizontal on the crate with a force . If u
is greater than some critical value, the woman
cannot start the crate moving no matter how
hard she pushes. This critical value of us is

Answer 1

Given:

Angle of force with the horizontal = 60°

Static friction = us

Kinetic friction = uk

To find:

The critical value of us.

Solution:

The normal force n= mg + F sin60°

The critical value of static friction so that the woman cannot move the crate, no matter how much force she applies. The acceleration will be 0, so the horizontal equation of motion is:

0 = F cosα - us*n

F cos 60° = us * (mg + F sin60°)

F cos 60° - us F sin 60° = us*mg

F = us * mg/ (cos 60° - us sin 60°)

For the critical value of us, the force should be greater then 0, i.e. nominator and denominator, both should be greater than 0.

cos 60° - us sin 60° ≥ 0

In the limit case, cos 60°- us sin60° = 0

us = cot 60° = 1/√3

Therefore, the value of us is 1/√3.