Truck deceleration to prevent box from sliding

1. Sep 22, 2009

Zhalfirin88

1. The problem statement, all variables and given/known data
The coefficent of static friction between the floor of a truck and a box resting on it is 0.38. The truck is traveling at 69.7 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide?

2. Relevant equations

3. The attempt at a solution
Not a clue, don't you need at least one mass?

2. Sep 22, 2009

PhanthomJay

Re: Friction

No. Try drawing a free body diagram of the box and note the forces acting on it. Then use Newton's laws.

3. Sep 22, 2009

Fightfish

Re: Friction

Basically the problem is about the maximum possible acceleration of the truck (to rest) such that the box remains in static equilibrium.
Consider the forces acting on the box when the truck is decelerating to rest. You will realise that the mass m of the box eventually cancels off to give a nice expression.

4. Sep 22, 2009

Zhalfirin88

Re: Friction

So you'd get

$$\mu$$s *mg = ma

$$\mu$$s *g = a

How does this help any?

5. Sep 22, 2009

CompuChip

Re: Friction

Doesn't that tell you the maximum acceleration that the box (hence, the truck) can have such that it won't start sliding?

6. Sep 22, 2009

Zhalfirin88

Re: Friction

Then what would you plug that into? I used:

vf2 - vo2 = 2a$$\Delta$$x

Which would give you:

$$\frac{-v_o^2}{2a}$$

But plugging in the numbers doesn't make sense. Because it'd look like: (velocity is in m/s)

$$\frac{-19.36^2}{2(3.4)}$$

7. Sep 22, 2009

PhanthomJay

Re: Friction

Since you are taking v_o as positive, the acceleration, in the opposite direction, is negative. That will give you a positive delta x. Also mu(g) is not 3.4, (.38(9.8) = 3.7).

8. Sep 22, 2009

Zhalfirin88

Re: Friction

Yes, I know, I did it earlier and I knew it was 3.something. But that wasn't what I was meaning, based off of that, the truck can stop in 50 m and the box will not slide?

9. Sep 23, 2009

PhanthomJay

Re: Friction

That looks about right. Just be sure to read the above posts to convince yourself why that is the correct solution.

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