A flatbed truck and Static Friction and Crate breaking loose

[shake94]

A flatbed truck and Static Friction and Crate "breaking loose"

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.

 

[TSny]

Hello.

For part a,
I got: y-dir:
Fnormal-Wsinθ=0

x-dir:
Wcosθ-Fsmax= 0

If θ in your expressions represents the 10^o angle of incline, then I think you should check to make sure you applied the trig correctly.

Also, just because there is a static friction force doesn't mean that the static friction force is at its maximum value.

 

[shake94]

Well, to show that the crate is not about to slip, I need to show that the component of gravity hasn't yet reached Fsmax. And that's my attempt here. And it breaks loose just when the component of gravity can no longer oppose it. So, how do I find θ value for when it breaks loose?

 

[TSny]

OK, I see now. I thought that θ in your expressions was the angle given in the problem and that you were going to work out what force of friction would be necessary to keep the crate from slipping on the flatbed at a 10^o angle.

Your approach will work fine, but I think you still need to fix the trig functions. To find θ you will want to substitute your expression for Fnormal from the y-direction equations into the expression for the forces in the x-direction and then simplify before trying to solve for θ.

 

[shake94]

I understand your poitn about the trig functions, I have understood them because they're a slope so I will have to change the angles as well. Thank you for confirming that.

Okay, so the new θ I find from substituting the equations that way will be the θ that is required to "break loose" the crate, correct?

Also, for the second part, please tell me if I am right. They are asking for the MINIMUM acceleration for the 10degree slope. So, here obviously velocity isn't constant and I will have to do the substitution part again but this time with θ= 10 degree and find acceleration that will break loose the crate. Is that right?
What can the driver do for the breaking loose to happen?

Sorry too many questions. :uhh:

 

[TSny]

Okay, so the new θ I find from substituting the equations that way will be the θ that is required to "break loose" the crate, correct?

That's right.

Also, for the second part, please tell me if I am right. They are asking for the MINIMUM acceleration for the 10degree slope. So, here obviously velocity isn't constant and I will have to do the substitution part again but this time with θ= 10 degree and find acceleration that will break loose the crate. Is that right?

Yes.

What can the driver do for the breaking loose to happen?

What can the driver do to make the velocity of the truck change?

 

[shake94]

He can change the direction, or accelerate. ?

 

[TSny]

Turning the steering wheel to change direction would be one possibility. There are also two different ways to produce acceleration along the x-direction.

 

[shake94]

Okay. I got it! Thank you so much! I really appreciate it!