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A flatbed truck and Static Friction and Crate "breaking loose"
A flatbed truck is going downhill on a slope of 10 degrees at a constant speed of 20 m/s (a little over 50mph). A crate of 500 kg is sitting on the flatbed , without being attached by ropes or cables. The flatbed and the crate experience friction , characterized by a static friction coefficient μs =0.35
a. Show that the crate is not about to slip and calculate the angle of the slope that would be required for the crate to "break loose"
b. For the 10 degree slope, calculate the minimum acceleration for which the crate is going to "break loose" and describe in words what the truck driver could do for this to happen (list all possible scenarios).
F=ma
V^2=u^2+2as
So I drew the FBD for the crate and made my coordinate system along the slope. I got W and Fnormal in the y-dir. and W and Fsmax in the x-dir. Since its going in constant velocity, acc would be 0.
For part a,
I got: y-dir:
Fnormal-Wsinθ=0
Fnormal= Wsinθ
x-dir:
Wcosθ-Fsmax= 0
Wcosθ-μsFnormal=0
θ=arccos(μsFnormal/W)
For part b,
Acceleration is involved so, I would do the same procedure but involve ma in the x-direction, correct?
Please help me on to go further from here.
I have a feeling I messed up somewhere with the angles because of the slop thingy. It sort of confuses me.
Homework Statement
A flatbed truck is going downhill on a slope of 10 degrees at a constant speed of 20 m/s (a little over 50mph). A crate of 500 kg is sitting on the flatbed , without being attached by ropes or cables. The flatbed and the crate experience friction , characterized by a static friction coefficient μs =0.35
a. Show that the crate is not about to slip and calculate the angle of the slope that would be required for the crate to "break loose"
b. For the 10 degree slope, calculate the minimum acceleration for which the crate is going to "break loose" and describe in words what the truck driver could do for this to happen (list all possible scenarios).
Homework Equations
F=ma
V^2=u^2+2as
The Attempt at a Solution
So I drew the FBD for the crate and made my coordinate system along the slope. I got W and Fnormal in the y-dir. and W and Fsmax in the x-dir. Since its going in constant velocity, acc would be 0.
For part a,
I got: y-dir:
Fnormal-Wsinθ=0
Fnormal= Wsinθ
x-dir:
Wcosθ-Fsmax= 0
Wcosθ-μsFnormal=0
θ=arccos(μsFnormal/W)
For part b,
Acceleration is involved so, I would do the same procedure but involve ma in the x-direction, correct?
Please help me on to go further from here.
I have a feeling I messed up somewhere with the angles because of the slop thingy. It sort of confuses me.
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