# Truck with stacked masses

1. Dec 10, 2015

### leaf

1. The problem statement, all variables and given/known data
Two crates of mass m are stacked on the back of a flat-bed truck of mass M. The truck accelerates from rest with a uniform acceleration a.
How does the force exerted by the road on the truck change when the masses are stationary vs. when the masses are sliding on the floor of the truck?

2. Relevant equations
Fn = (2m+M)g
Ff = μs(2m+M)g
as = μg (as is value of acceleration beyond which crates begin sliding, and coefficient of static friction between crate and truck is μ)

3. The attempt at a solution
The first two equations above are the equations I got for the force exerted by the road when the masses are stationary, but I'm not sure how this would change when the masses are sliding. It doesn't seem to me that they would, but maybe I'm missing something. The part of the question asking for force exerted by the road when the masses are sliding comes after a question about the value of as so I'm thinking that maybe that has something to do with it?

2. Dec 10, 2015

### Staff: Mentor

If they slide, their acceleration is not "a" any more. What does that mean for the horizontal force between the lower mass and the truck?

3. Dec 10, 2015

### BvU

Your second equation has a $\mu_s$ and (since it is $\mu_s F_N$ for truck + crates), it means $F_f$ is the maximum friction force between road and truck. $\mu_s$ is a coefficient that has to do with the tyres and the road surface, not with the crates! If the truck accelerates with less than $m_s g$ the actual friction force is less than $F_f$ !

$\mu$ in your third equation is what ?
If you really mean $\mu_s$ then it seems a coincidence that your as is indeed the maximum acceleration the crates can have without sliding !

Let me -- for clarity -- use $\mu_c$ for the friction coefficient between crates and truck bed.

When the crates are sliding, they can't exercise more than the full $2 \mu_c mg$ on the truck bed, so they don't accelerate with a but with a smaller acceleration $\mu_c g$ . The road still hat to provide all accelerating force, but for the truck it is a and for the crates it is less than a.

4. Dec 10, 2015

### leaf

Thanks for the help, but I'm still a bit stuck.

BvU, μ in the third equation is the static friction between the crates and the truck and μs is the static friction between tires and road.

The friction force between truck and crates would decrease to μmg when the crates begin sliding I believe ($\frac{1}{2}$μ is the coefficient of kinetic friction between truck and crates given in the problem and the combined masses of stacked crates is 2m), but I'm still confused about how this relates to the force between truck and road.

5. Dec 10, 2015

### BvU

I agree that $a_s = \mu g$ is indeed the maximum acceleration the truck can give the crates. And that maximum acceleration requires a force $F_{\rm truck\rightarrow crates} =2 \mu mg$. With a simple free body diagram of the truck loaded with the sliding crates and accelerating with an acceleration $a > a_s$ you can now symbolically write down the force the road has to exercise on the truck.

If the coefficient of kinetic friction is lower than the $\mu$ we used for static friction, then that $\mu_k$ takes the place of the static $\mu$.
But it can well be that your answer doesn't have to be all that specific and all that is wanted is a qualitative answer ('smaller', 'bigger', ...)