# Friction Force Help

1. Feb 23, 2006

Hello everyone,

I have have no clue on how to go about this problem here, if someone can explain how to go about these it would be great! THANK YOU ALL!

Problem 1: :grumpy:

A 62 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal.
(a) If the coefficient of static friction is 0.51, what minimum force magnitude is required from the rope to start the crate moving?
(b) If µk = 0.36, what is the magnitude of the initial acceleration of the crate?
m/s2

Last edited: Feb 23, 2006
2. Feb 23, 2006

Ok so i found a) by using T = mg*u / cos x + sin x * u

yielding => 282.23 Newtons,

trying to find B) can anyone help?

EDIT: Just a little help to get me moving is good enough, PLEASE SOMEONE HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Last edited: Feb 23, 2006
3. Feb 23, 2006

still no post ?

Last edited: Feb 23, 2006
4. Feb 23, 2006

So no ones gonna help me, thanks alot

5. Feb 23, 2006

### Cyrus

I think your part (a) is wrong.

Where did you get this?

T = mg*u / cos x + sin x * u

6. Feb 24, 2006

No question a is correct i already put it in webassign, (a online physics homework thing for college). for b i just asked my professor, it was pretty simple. Using F = ma
F = Fnet while in motion

Thanks anyway dude

7. Feb 24, 2006

### topsquark

It strikes me as somewhat odd that you found the answer to a), but can't get b). Can I assume the T equation was something from your book? (Otherwise I can't figure out how you got it!)

b) The object has a normal force (N) and a weight (w) in the vertical direction. I took the applied force (F) to be acting up and to the right, so the friction force (f) will be acting to the left. I choose a coordinate system with +x to the right and +y upward.

As is typical with a friction problem, you will need to do the net force in the y direction, so do that first:
$$\sum F_y=Fsin \theta +N-w=0$$
So you can find N.
In the x direction we note that the box is presumed to be sliding, so we are using kinetic friction. $$f_k= \mu_kN$$ so we have:
$$\sum F_x=Fcos \theta-f_k=ma$$.

Plug in for kinetic friction and the normal force and solve for a.

To do a) you basically do the same thing. The difference is that we are looking at a stationary situation, so we use static friction. Since we are looking for the minimum force required, we are using maximum static friction and a=0 in the limit. This will allow you to derive the equation you used for part a).

-Dan