# Static and Kinetic Friction. Box sliding in the back of a truck.

1. Feb 5, 2007

### robbondo

1. The problem statement, all variables and given/known data
Losing Cargo. A box of mass m rests on the flat floor of a truck. The coefficients of friction between the box and floor are "us" and "uk". The truck stops at a stop sign and then starts to move with an acceleration of a. If the box is a distance x from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck? How far does the truck travel in this time? Take the vertical acceleration to be g.

2. Relevant equations

3. The attempt at a solution

Well, I'm pretty stuck here. I know that there must be some force greater than "us"n to get the box to start moving. And then it will accelerate f=ma with the force being the force of the trcuk moving - the force "uk"n. So, where I'm stuck is relating these values to time. How long does it take the box to become unstuck and overcome the static friction? I'm super confused. I know if I get the first part how to solve the second. Any help is greatly appreciated.

2. Feb 5, 2007

### denverdoc

The question I think is whether the box becomes unstuck or not, not how much time it takes. And with the given data we dont know for sure? Its whether:

a>us*g,
if it is, then I believe you should assume it begins to slide the instant the truck moves, in other words forget the trucks motion for the first part of the problem, and treat the box as if force of Ma is applied in the direction of the tailgate.

3. Feb 7, 2007

### robbondo

Well, so I've figured out that the static friction shouldn't be a factor in this problem because if it's gonna move, it should start sliding immediatly, so with that information I did the following.

Where I think that I went wrong was using a for the acceleration of the box. So if the acceleration of the truck=acceleration then.

$$\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}$$

so the net force acting on the box is

ma - "uk"mg=F(in the x dir.)

do if the net force acting on the box is $$\ ma-u_kmg / m = a$$

then by pluggin that acceleration of a-"uk"g into the distance formula

$$x = x_0 + v_0 t + (1/2) a t^2$$

I get that t = the square root of (x/(1/2(a-("uk"g))) but this answer is wrong. I think that the main mistake in my reasoning is assuming that the acceleration of the box is equal to the acceleration of the truck. I can't figure out what the force of the truck on the box is... If I can just get either one of those two things I think I should be ok.

4. Feb 7, 2007

### denverdoc

lost me there, i get x=1/2*(a-gu)*t^2, solving for t;
2(x)/(a-gu)=t^2

Maybe its just your convention of keeping the 1/2 under the x. It was my idea to swap inerttial reference frames. What is the answer?

Last edited: Feb 7, 2007
5. Feb 7, 2007

### robbondo

yeah ur right. I messed up on the algebra. Thanks.

6. Feb 7, 2007

### denverdoc

sigh of relief. Good problem.

7. Feb 7, 2007

### PhanthomJay

That should give you the correct answer, however, I don't understand your reasoning in determining the net force acting on the block. There is only one force acting on the block horizontally, and that is the friction force, u_k(mg). Applying Newton 2 in a FBD of the block
F_net = m(a_block)
u_k(mg) = m(a_block)
a_block = u_k(mg) with respect to the ground, acting in the same direction as the truck's acceleration.
The acceleration of the block with respect to the truck is then
a_block w/respect to truck = (a -u_k(mg)).

So this yields the same result, but you should recognize that the block's acceleration and the truck's acceleration are different, and Newton 2 often fails you when you look at the entire system when there are relative movements involved.

8. Feb 7, 2007

### denverdoc

agreed. It was my suggestion, not the OP's to use this tactic. Should probably just laid it out conventionally as you have done.