# A packing crate.....

• Jun 29th 2006, 09:36 AM
pocahontas
A packing crate.....
A packing crate is placed on a plane inclined at an angle of 35o from horizontal. If the coefficient of static friction between the crate and the plane is 0.65, will the crate slide down the plane?
• Jun 30th 2006, 05:33 AM
topsquark
Quote:

Originally Posted by pocahontas
A packing crate is placed on a plane inclined at an angle of 35o from horizontal. If the coefficient of static friction between the crate and the plane is 0.65, will the crate slide down the plane?

I can't do diagrams. Sorry!

A quick description of how I'm visualizing this: I am taking the slope to be from top left to bottom right ( the direction \ ).

Sketch a free body diagram. Assuming the crate to be stationary there are three forces: the weight, the normal force on the crate, and the static friction force. The condition for the crate to be stationary is that the weight component down the incline is less than the maximum possible static friction force.

The weight component down the incline will be mgsin(theta). (NOT cosine! Be sure you understand why.) The maximum static friction force is defined to be $\displaystyle \mu_sN$. In this case N = mgcos(theta). (Do a Newton's second law in the direction of the Normal to get this.) So the maximum static friction force will be $\displaystyle \mu_s mgcos(\theta)$.

You don't have a mass given. So you can do two things:
1) Assume a value for m. When in doubt, m = 10 kg.
2) Subtract the weight component from the maximum static friction force. If the crate is stationary, this value will be positive. So:
$\displaystyle \mu_s mgcos(\theta) - mgsin(\theta) = mg \left (\mu_s cos(\theta) - sin(\theta) \right )$
If the value in parenthesis is positive, the crate will be stationary.

-Dan