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Truck carrying a box, friction problem

  • Thread starter GrF58
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Homework Statement



A truck (mass M) is carrying a box (mass m) in its flatbed, traveling along a straight, level road at speed v, when the driver sees a deer in the road a distance d ahead. The driver slams on the brakes to cause the truck to skid to a stop. The coefficient of kinetic friction between the road and the tires is μt, the coefficient of static friction between the box and the truck bed is μs, and the coefficient of kinetic friction between the box and the truck bed is μk. Draw the free body diagrams of both the box and the truck during the deceleration, and list of all Third Law pairs between the truck and the box. Give Newton's second law equations for the box and the truck, in each direction (4 equations). Does the box slip off the truck? Does the truck hit the deer?

Homework Equations



Fnet = ma
Force of friction = μN

The Attempt at a Solution



Now this is a symbolic question (my teacher did not provide numbers at all for it), but I'm stuck with my free body diagrams. I have:

Box:
- Gravity pulling down (mg)
- Normal force pushing up from the truck (equal and opposite to mg)
- Some form of friction, but I have no idea. The problem does not specify if it's skidding or staying still. If static, it should be a force pointing in the direction of motion, if kinetic, it should be a force pointing in the opposite direction, yes? But I don't know which to use in my FBD.

Truck:
- Gravity pulling down (Mg)
- Weight of the box pushing down (mg)
- Normal force up from the road (equal and opposite to Mg + mg)
- Kinetic friction, opposite the direction of motion (μt)
- I'm not sure if the friction from the truck on the box is also applying a force on the truck, or what that force might even be.
 

Answers and Replies

  • #2
haruspex
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If static, it should be a force pointing in the direction of motion,
Wouldn't that make the box go even faster?
not sure if the friction from the truck on the box is also applying a force on the truck,
One of Newton's laws answers that.
 
  • #3
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Wouldn't that make the box go even faster?
You’re right. Static friction try to keep the box still with respect to the truck, so it will face the same direction as acceleration (opposite the motion), correct?

One of Newton's laws answers that.
Third law, I’m assuming. So the static friction from the box will exert the same force on the truck in the opposite direction, facing in the same direction as the motion? I’m still confused where the kinetic friction between the box and truck will come into play.
 
  • #4
haruspex
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Static friction try to keep the box still with respect to the truck, so it will face the same direction as acceleration (opposite the motion), correct?
Yes.
So the static friction from the box will exert the same force on the truck in the opposite direction, facing in the same direction as the motion?
Yes.
I’m still confused where the kinetic friction between the box and truck will come into play.
What will be the truck's deceleration? Might the box slip?
 
  • #5
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What will be the truck's deceleration? Might the box slip?
I think the box will slip when the acceleration > μsN/m.

Would the kinetic friction force on the box just be facing in the same direction as the static friction force (opposing motion)? But it feels like there would be no forces from kinetic friction until/if the box began to slide, and I'm not sure how to represent that on an FBD.
 
  • #6
haruspex
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Would the kinetic friction force on the box just be facing in the same direction as the static friction force (opposing motion)?
Yes, but to be clear, friction opposes relative motion of the surfaces in contact. In the present case, that happens to mean it opposes the motion relative to the ground of the box, but is the other way on the truck.
I'm not sure how to represent that on an FBD.
You need to consider the two cases separately. What relationship between the given variables determines whether it will slide?
 
  • #7
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So I spoke with my professor, and she basically told me that both static and kinetic friction between the truck and the box can effectively be represented using the same force on the FBD, as long as I note it as such. And since we're only concerned with whether or not the truck will hit the deer, and whether or not the box will start to slide (and not what happens while it slides), can we effectively ignore it for this problem?

My 2nd Law equations ended up being: (where ax & ay is acceleration in the x and y directions, respectively)
Sum of F in the x direction on the truck = M*ax = force from friction on truck from box - force from kinetic friction on tires
Sum of F in the y direction on the truck = M*ay = 0 = normal force from road on truck - normal force from box on truck - Mg

Sum of F in the x direction on the box = m*ax = -force from friction on box from truck
Sum of F in the y direction on the box = m*ay = 0 = normal force on box from truck - mg

And the 3rd law pairs are:
normal force on the box from the truck = the normal force from the truck on the box
friction force on the box from the truck = the friction force from the truck on the box

Now to find if the truck is going to hit the deer, I was going to use the formula: vf^2 = v0^2 +2ad
To find the distance d the truck will travel.
And solve for d = -v^2/2a (since vf = 0)

But I still need acceleration, so now I'm wanting to find the acceleration of the truck in the x direction (which I was going to do by a = F/M, then using the x directional forces on the truck as = F).

I've gotten to this point, but I want to be sure I'm not making some grievous mistake before I continue.
 
  • #8
haruspex
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Sum of F in the x direction on the box = m*ax
Same ax as for the truck? So you are assuming the box does not slide? If not, you need to use a different variable.

I see no equations involving the friction coefficients.
 
  • #9
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Same ax as for the truck? So you are assuming the box does not slide? If not, you need to use a different variable.

I see no equations involving the friction coefficients.
My apologies, I should have been more clear.

When I say the force from kinetic friction on tires for example, I'm referring to the equation: μt * Normal force from the road on the truck

Same with acceleration, I wrote them down as atx, aty, abx, and aby, but I was being lazy with typing the notation I used. I was just hoping I was on the right track.

I did get further along with it, and tried to find the acceleration of the truck using a = F/m, so:

atx = [μs*mg - μtires*(Mg + mg)] / M

Essentially, I'm trying to find the atx by adding up all the forces applied to the truck in the x direction (which are only two: the friction between the truck and the road, and the friction between the truck and the box), then dividing by the mass of the truck. Though I could be totally off base.
 
  • #10
haruspex
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atx = [μs*mg - μtires*(Mg + mg)] / M
Yes, if the box does not slide.
Can you find the condition for whether the box slides?
 
  • #11
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The acceleration of the truck if the box is sliding would replace μs with μk in the formula.

As for whether the box will slide, the max force of static friction = μs*N = μs*mg

Plugging that into a = F/m, the maximum acceleration the box can handle without slipping is μs*g.

IF the box is not sliding, then abx = atx.

So, as long at atx <= μs*g, then the box should not slide, right?
 

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