Friction rope tied to crate

A rope is tied to a large crate, which is sitting on a flat surface. The coefficient of static friction between the crate and the ground is 0.9. If a person is to pull on the rope with the minimum force needed such that the crate begins to slide, the angle between the rope and the ground should be
A)

greater than 0 degrees but less than 90 degrees
B)

0 degrees (rope is horizontal)
C)

90 degrees

I know the answer is A. I found this question in practice problems for midterms.

Now, I was trying to prove it:
horizontal pulling: F=ukmg.

greater than 0 degrees but less than 90 degrees:
Horizontal component: uk(mg-sinthetaF) (Let F be magnitude of force.)
Vertical component: costhetaF
Total magnitude:
(Pythagoras) and (sin^2theta+cos^2theta=1)
0.81mg-1.8sinthetaF+F^2=Magnitude.

How do I prove 0.81mg-1.8sinthetaF+F^2 > 0.9mg?

http://ift.tt/1kJ3C6L

Leave a comment

Your email address will not be published.


*


Show Buttons
Hide Buttons