# Friction between a crate and an accelerating truck

1. Mar 28, 2009

### tascja

1. The problem statement, all variables and given/known data
you and your friend have just loaded a 200kg crate filled with priceless art object into the back of a 2000kg truck. As you press down on accelerator, a force, Ft, propels the truck forward. What is the max magnitude the force can have without the crate sliding? The static and kinetic coefficient of friction between the crate and the bed of the truck are 0.8 and 0.3.

2. Relevant equations

3. The attempt at a solution
Here is what i have for my FBD
The crate:
>normal force acting up
>gravitational force acting down
>static force acting to the right

The truck:
> Ft acting to the right
>gravitational force acting down (includes both the weight of the crate and the truck)
>normal for acting up

I think that as the truck accelerates to the right the crate would want to slip to the left, and the friction would oppose this motion (acting to the right) and stops the sliding. but can someone explain how i can connect the two FBD to relate Ft to the friction force?

2. Mar 28, 2009

### Staff: Mentor

Good.

OK, but strictly speaking the weight of the crate is transmitted to the truck as a downward normal force. (The gravitational force of the crate acts only on the crate.)
Hint: What's the maximum force that the truck can exert on the crate without slipping? What acceleration would that imply?

3. Mar 28, 2009

### tascja

ok so i decided to look at this as just two boxes on top of one another...

so on the upper box the only force being applied in the x direction is the friction
so Fnet = ma
ma = μFn
(200)a = (0.8)(1960)
a = 7.84 m/s^2

so im not sure if i can assume that both boxes then have the same acceleration? but if i can then:
For the lower box:
Fnet = ma
= (2000)(7.84)
= 15680 N
therefore the max magnitude of Ft = 15680 N

4. Mar 28, 2009

### Staff: Mentor

Good!

Well.... if they don't slip, they must move together.

That's the net force on the truck. Don't forget that the crate also exerts a backward friction force on the truck. (I forgot to point that out on your FBD for the truck--you were missing that force.)

An even easier way to view it is to now treat "truck + crate" as a single system. What force is required to give it the needed acceleration?

5. Mar 28, 2009

### tascja

is just including both their masses in the Fnet sufficient?

6. Mar 28, 2009

### Staff: Mentor

Yes. Assuming you mean Fnet = (m1 + m2)*a = Applied force Ft.

7. Mar 28, 2009

### tascja

thank you for all your help Doc Al!! =)