# Frictional force problem

1. Aug 11, 2010

### SugerQueen

In unloading a truck a man allows a 120kg crate to slide down a plank inclined at 30.0o to the horizontal. The crate accelerates at 3.90ms-2. What is the size of the frictional force?

I know I need an equation for frictional force on an incline plane to slove this but thats where my problem lies I didn't learnt it in class and I can't find the equation.
my guess is: Ff= umgcosa
but if it is i'm also stuck on how to find u in this problem?

u=F/N but then how do i find N? and F? I'm really stuck and going in circles.
Hope my ramble makes sense to someone lol Thank you for any help

2. Aug 11, 2010

### N-Gin

Draw the free body diagram. If the crate accelerates at $\vec{a}$ you have

$$m\vec{g}+\vec{N}+\vec{F_{f}}=m\vec{a},$$

where $\vec{N}$ is the normal force prependicular to the incline. What is it equal to and why?

3. Aug 11, 2010

### SugerQueen

ok so if i put that all together i get an answer to my original problem of 3971.9N? does that sound right? or am i so far off its not funny?
Thank you :) so confused right now lol

4. Aug 11, 2010

### N-Gin

Decompose the vectors in x and y directions, where x is parallel with the incline and y perpendicular to it. Your answer is greater than the weight of the crate. It means that the crate goes up the incline, not down as it should.

Maybe this can help

x-direction: $$ma=mg\sin \alpha - F_{f}$$

y-direction: $$N=mg\cos \alpha$$

5. Aug 11, 2010

### SugerQueen

So I was way off. :)
I re-did after your last reply and got:
120*3.9=120*9.8*sin30-Ff
468=588-Ff
Ff=120N
N=1,018.45
I'm not sure if this then goes into the u equation but thats what i did:
u=120/1,018.45
u=0.1178
or does the answer above Ff=120N the overall answer to the question?
Sorry to be such a bother but i just can't get my head around this. It's a mind block.

6. Aug 11, 2010

### N-Gin

Yes, I got the same result.

7. Aug 12, 2010

### SugerQueen

ok thank you :D for all your help :)