Coefficient of kinetic friction for 300 kg crate on 25° incline

[studentmom]

how to find coefficient of kinetic friction... please?!

I am lost on this question, hopefully someone can help:

a 300 kg crate is placed on an adjustable incline and as one end of the incline is raised the crate begins to move downward; if the crate slides down the plane at an acceleration of 0.7m/s2 when the incline angle is 25 degrees, what is the coefficient of kinetic friction between the ramp and the crate?

I drew a free-body diagram and that's about as far as I got... I know that the kinetic friction force is in the opposite direction that the crate is moving, but other than that I'm lost... please help!

Thanks!

 

[humaders99]

do u have any lengths? Cause if you do you can just us tanθ and that would give you the answer...im doing a similar question and i figured out through other equations that tanθ is how u get it. So if the length of the incline was 4m and the height was 3 then ur answer would be 3/4.

 

[studentmom]

no, the problem is stated as is, no lengths given...

 

[BishopUser]

First thing i would do is draw a free body diagram, and create an coordinate system in which the x-axis is parallel to the plane. There are three forces acting on the box: gravity, friction, and the normal force. Friction = (coefficient of kinetic friction*normal force). On our coordinate system we know that the sum of the forces in the y direction must be 0 (the box isn't going to jump off the plane or anything like that). since friction does not have a component in the y-direction we can ignore it for now. The first step is to calculate the normal force.

Fy = -mgy + normal force
0 = -(300kg)(9.8m/s^s)(cos25) + Fn
Fn = 2664.5N

now that we have the normal force we can use F=ma in the X-direction. We have 2 forces with components in the x-direction: gravity, friction

Fnet = MA
Fgravity - Ffriction = MA
(9.8 m/s^2)(300kg)(sin25) - (2664.5N)(X) = (300kg)(.7m/s^2)
X = .39 (the coefficient of kinetic friction)

 

[studentmom]

Thank you so much for your help!

 

[Razza]

I know this is an old thread and I know probably no ones going to see it, but where did the force of friction (2664.5N) come from? Seeing as 9.8*300 doesn't equal that, and 300*0.7 certainly doesnt...

 

[alphysicist]

Hi Razza,

2664.5 is not the force of friction in that post; it is the normal force they calculate in the first three equations of post #4. In the equations after that they show that the magnitude of the force of friction is (2664.5)(X), where X is the coefficient of kinetic friction that they are solving for.

 

[Kushal]

*edit* ooops alphysicist, i think we posted at the same time.the force of friction is not 2664.5 N. It is (2664.5)(X) N, where 2664.5 N is the normal contact force and X is the coefficient of friction.

frictional force = normal contact force x coefficient of friction

 

[Razza]

Thank you guys,

Haha its right above it how could I have missed. I think that was just because I was doing my own equation at the same time using my variables and that's where it got me lost.

Thank you for answering such an obvious question though.

 

[Razza]

Okayy wow..


With this, my equation works out to be:

F= -mgy+ Fn
0=-(0.012 kg)*(9.8 m/s squared) +Fn
-Fn = -0.1176 N


Then to sub it in

Fnet = MA
Fgravity - Ffriction = MA
(9.8 m/s^2)(0.012kg)(sin61) - (0.1176 N) (X) = (0.012kg)(15.24m/s^2) (as long as velocity is two times the average speed)

= 0.168

Now that sounds a lot better.

The answer I got given another formula, gave me 1.8, which seemed too high and gave my Force an answer of roughly 211 instead of 19 like it did for this.

F=0.168*117.6(my normal force)

Is that correct?

 

[alphysicist]

Okayy wow..


With this, my equation works out to be:

F= -mgy+ Fn
0=-(0.012 kg)*(9.8 m/s squared) +Fn

What is the statement of the problem? If it's the same type of problem as the original problem in this thread (object moving down an inclined plane) then I believe you are missing a trig factor here. As the angle of the incline increases, the normal force should decrease.

-Fn = -0.1176 N


Then to sub it in

Fnet = MA
Fgravity - Ffriction = MA
(9.8 m/s^2)(0.012kg)(sin61) - (0.1176 N) (X) = (0.012kg)(15.24m/s^2) (as long as velocity is two times the average speed)

Here again, it's hard for me to be sure what's going on without the full problem. However, if this was just a normal incline problem with the only forces being from gravity and the incline, this acceleration seems too high. If the object was dropping freely, the acceleration would be 9.8 m/s^2. Here on an incline means the acceleration by gravity down the incline is not as much, and in addition you have friction to slow it down even more.