# Static friction

1. Sep 20, 2009

### semc

You try to push a crate of weight M with a force F on a horizontal floor. The
coefficient of static friction is μs, and you exert the force F under an angle φ below the
horizontal.
Determine the minimum value of F that will move the crate.

Okay so basically what i did was equate Mμs=Fcos φ and evaluate for F. however the answer my teacher gave me is slightly different. He use Nμs=Fcos φand he found N by evaluating the forces in the y component. So the question is why cant i just equate the force applied with the static friction?

2. Sep 20, 2009

### pgardn

Does your M stand for mass? If so your units will come out wrong for force.
If one pushes (y-component) down on a mass sitting on a surface then the Normal force is going to be different compared to the same mass pushed on perfectly horizontally, or the mass pushed up. So in your problem you are pushing horizontally and vertically.

3. Sep 20, 2009

### RoyalCat

His M is weight, not mass, so unit-wise it works.

But the problem is just as you've pointed out.

The definition of the kinetic frictional force is: $$f_{s max}=\mu_s N$$ and NOT: $$f_{s max}\neq \mu_s W$$

By pushing up on the mass, you're decreasing the normal force from the ground, thereby decreasing the friction force, and making life easier for yourself.

Last edited: Sep 20, 2009
4. Sep 20, 2009

### pgardn

Oops my bad as they say.

That is a really a potentially confusing symbol to use for weight...