Calculating Frictional Force on an Inclined Plane

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Homework Help Overview

The problem involves calculating the frictional force acting on a crate sliding down an inclined plane. The crate has a mass of 120 kg and is on a plank inclined at 30 degrees, with an acceleration of 3.90 m/s². The original poster expresses confusion about the relevant equations and how to determine the coefficient of friction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need for a free body diagram and the relationship between forces acting on the crate. There are attempts to derive equations for friction and normal force, with questions about how to find necessary variables like the coefficient of friction.

Discussion Status

Some participants have provided guidance on breaking down the forces into components and checking calculations. There is acknowledgment of confusion regarding the results, with some participants questioning the validity of their answers and interpretations of the problem setup.

Contextual Notes

The original poster mentions not having learned the relevant equations in class, which may limit their ability to solve the problem independently. There is also a concern about the physical interpretation of the results, particularly regarding the direction of the crate's motion.

SugerQueen
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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 that's where my problem lies I didn't learned 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
 
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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?
 
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
 
SugerQueen said:
ok so if i put that all together i get an answer to my original problem of 3971.9N? does that sound right?

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
 
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 that's 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.
 
SugerQueen said:
or does the answer above Ff=120N the overall answer to the question?

Yes, I got the same result.
 
ok thank you :D for all your help :)
 

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