Truck deceleration to prevent box from sliding

[Zhalfirin88]

Homework Statement

The coefficient 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?

Homework Equations

The Attempt at a Solution

Not a clue, don't you need at least one mass?

 

[PhanthomJay]

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

 

[Fightfish]

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 realize that the mass m of the box eventually cancels off to give a nice expression.

 

[Zhalfirin88]

So you'd get

$$\mu$$s *mg = ma

$$\mu$$s *g = a

How does this help any?

 

[CompuChip]

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

 

[Zhalfirin88]

Then what would you plug that into? I used:

v_f² - v_o² = 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)

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

 

[PhanthomJay]

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).

 

[Zhalfirin88]

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).

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?

 

[PhanthomJay]

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