Box moving on a decelerating truck?

[dtseng96]

Homework Statement

A truck with a steel bed carries a wooden box, where μ_s = 2μ_k. The box of mass m is in the middle of the bed of length l. If the truck is moving forward with velocity v_o but is slowing down, determine the maximum distance the truck can take to stop before the box hits the cabin of the truck in terms of the variables given.

Homework Equations

F = ma, F_k = μ_kmg

The Attempt at a Solution

So my teacher said that the acceleration of the box relative to the truck is a_B = μ_kg, while the acceleration of the truck relative to the ground is a_T = -μ_kg. I'm having trouble understanding how my teacher got this... I do know that the acceleration of the block relative to the ground is a = μg, but I don't know how my teacher got the acceleration of the block relative to the truck. Can anyone help clarify my teacher's reasoning?


Any help is appreciated. Thank you so much!

In case you guys want to look at my teacher's actual solution, here it is. My question is about part b): https://dl.dropbox.com/u/12682457/m%20%26%20f%20%232%20copy.jpg

 

[Spinnor]

You wrote,

"...determine the maximum distance the truck can take to stop before the box hits the cabin..."

Shouldn't that be the minimum distance the truck can take to stop? If we decelerate slowly the box won't move relative to the truck.

Edit, ignore the above question, it now makes sense to me.

 

[Spinnor]

Does this look right?

 

[dtseng96]

@Spinnor: Yes, that does look correct to me! :) Except the question is looking for the maximum distance the truck travels :O

I discussed this again with my teacher, and apparently, there might be a mistake in my teacher's work. But yes, thank you for your solution! It gave me an idea of what the correct solution should look like...

 

[Nickie]

I would approach this problem by breaking it down into smaller parts and analyzing the forces acting on the box and the truck separately. First, let's consider the forces acting on the box.

The box is experiencing two forces: the force of friction (Fk) and the force of gravity (mg). The force of friction can be broken down into two components: the static friction force (Fs) and the kinetic friction force (Fk). Since the box is not moving relative to the truck, the static friction force must be equal in magnitude to the force of gravity (Fs = mg). This means that the box is not accelerating relative to the truck (aB = 0).

Now, let's consider the forces acting on the truck. The truck is experiencing the force of friction from the road (Ff) and the force of the box pushing against it (Fb). Since the truck is decelerating, the force of friction must be greater than the force of the box (Ff > Fb). This means that the truck is accelerating in the negative direction (aT = -Ff/m).

To determine the maximum distance the truck can take to stop before the box hits the cabin, we need to consider the distance the truck travels while decelerating (d) and the distance the box travels while the truck is decelerating (d'). Since the box is not accelerating relative to the truck, its distance traveled while the truck is decelerating must be equal to the distance traveled by the truck (d' = d).

We can use the equation d = vo*t + (1/2)*aT*t^2 to solve for the maximum distance (d) in terms of the given variables. Plugging in the values for aT and solving for t, we get t = vo/μg. Substituting this into the equation for d, we get d = (vo^2)/(2μg). This is the maximum distance the truck can take to stop before the box hits the cabin.

In summary, the reasoning for the acceleration of the box relative to the truck (aB = 0) and the acceleration of the truck relative to the ground (aT = -Ff/m) is based on the fact that the box is not moving relative to the truck and the truck is decelerating. I hope this helps to clarify your teacher's reasoning.