Friction between a crate and an accelerating truck

[tascja]

Homework Statement

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.

Homework Equations

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?

 

[Doc Al]

Here is what i have for my FBD
The crate:
>normal force acting up
>gravitational force acting down
>static force acting to the right

Good.

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

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.)

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?

Hint: What's the maximum force that the truck can exert on the crate without slipping? What acceleration would that imply?

 

[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 I am 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

 

[Doc Al]

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

Good!

so I am not sure if i can assume that both boxes then have the same acceleration?

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

but if i can then:
For the lower box:
Fnet = ma
= (2000)(7.84)
= 15680 N
therefore the max magnitude of Ft = 15680 N

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?

 

[tascja]

What force is required to give it the needed acceleration?

is just including both their masses in the Fnet sufficient?

 

[Doc Al]

is just including both their masses in the Fnet sufficient?

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

 

[tascja]

thank you for all your help Doc Al! =)