Calculating Frictional Force on an Inclined Plane

In summary, a man unloads a 120kg crate by sliding it down a plank inclined at 30.0o to the horizontal. The crate accelerates at 3.90ms-2 and the man is trying to find the size of the frictional force. Using the equation for frictional force on an inclined plane, Ff= umgcosa, the man gets stuck on finding the coefficient of friction (u). By drawing a free body diagram and decomposing the vectors in the x and y directions, the normal force (N) is found to be equal to 1,018.45N. After re-doing the calculations, the man arrives at the correct answer of Ff=120N.
  • #1
SugerQueen
15
0
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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  • #2
Draw the free body diagram. If the crate accelerates at [itex]\vec{a}[/itex] you have

[tex]m\vec{g}+\vec{N}+\vec{F_{f}}=m\vec{a},[/tex]

where [itex]\vec{N}[/itex] is the normal force prependicular to the incline. What is it equal to and why?
 
  • #3
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
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: [tex]ma=mg\sin \alpha - F_{f}[/tex]

y-direction: [tex]N=mg\cos \alpha[/tex]
 
  • #5
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.
 
  • #6
SugerQueen said:
or does the answer above Ff=120N the overall answer to the question?

Yes, I got the same result.
 
  • #7
ok thank you :D for all your help :)
 

1. What is frictional force?

Frictional force is the resistance force that occurs when two objects are in contact and moving relative to each other. It is caused by the microscopic irregularities on the surfaces of the objects, which create a force that opposes motion.

2. How is frictional force calculated?

Frictional force can be calculated using the equation F = μN, where F is the frictional force, μ is the coefficient of friction, and N is the normal force (the force perpendicular to the surface of contact).

3. What factors affect the magnitude of frictional force?

The magnitude of frictional force is affected by the type of surfaces in contact, the coefficient of friction, and the normal force. Rougher surfaces, higher coefficients of friction, and higher normal forces will result in greater frictional forces.

4. How does frictional force affect motion?

Frictional force opposes motion, so it can either slow down or stop an object's movement. It also causes objects to heat up due to the energy lost during the frictional process.

5. How can frictional force be reduced?

Frictional force can be reduced by using lubricants, such as oil or grease, between the two surfaces in contact. Changing the type of surfaces or decreasing the normal force can also reduce frictional force.

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