Newton's second law -- Crate sliding in the back of an accelerating truck

[Franklie001]

Ok so do i need to consider the mass of the crate on the truck as whole so (mass of the truck + mass of the crate) what about the air forces (Fa1 and Fa2) do i need to add them too, or are they considered as internal forces?

 

[PeroK]

Ok so do i need to consider the mass of the crate on the truck as whole so (mass of the truck + mass of the crate) what about the air forces (Fa1 and Fa2) do i need to add them too, or are they considered as internal forces?

Here's a clue:

 

[Franklie001]

Thanks

 

[Steve4Physics]

@Franklie001, if you want to do it using simultaneous equations, here are the steps...

’1’ is the truck and ‘2’ is the crate.

1. Find the resultant force on the truck, $F_1$, the sum of all the forces on the truck.
Hey, I'll even do that one for you:
Reading-off the truck's free body diagram gives:
$F_1 =5000 - 600 - F_{friction} = 4400 - F_{friction}$.

2. Find the resultant force on the crate, $F_2$, the sum of all the forces on the crate.
You do that one yourself.

3. Apply $F_1=m_1a$ to the truck to get equation 1.
I'll do that one for you:
$4400 - F_{friction} = 8000a$ (equation 1)

4. Apply $F_2=m_2a$ to the crate to get equation 2.
You do that one yourself.

5. You now have 2 simultaneous equations with unknowns $a$ and $F_{friction}$. You solve them.

If you don't like simultaneous equations, see the suggested method in my previous posts.

 

[Franklie001]

I've used the simultaneous equations and i 've got for the acceleration ax= 0.39 m/s^2 and for the force of friction f= -880N
Is that right?

 

[PeroK]

I've used the simultaneous equations and i 've got for the acceleration ax= 0.39 m/s^2 and for the force of friction f= -880N
Is that right?

Sadly that doesn't look quite right. Did you remember the air resistance on the crate?

 

[Franklie001]

No you are right I've got for the acceleration ax=0.382 m/s^2
and for the friction force f = 944N

Am i right?

 

[Franklie001]

So if the friction force is 944N i have an equal and opposite friction force acting on the crate.
Therefore the second solution is given already Ff=944N.
And the coefficient is 0.043

 

[PeroK]

So if the friction force is 944N i have an equal and opposite friction force acting on the crate.
Therefore the second solution is given already Ff=944N.
And the coefficient is 0.043

I get a different answer. Did you remember the air resistance on the crate?

 

[Franklie001]

I've got Friction force = 1040N
and coefficient of kinetic friction u = 0.048

 

[Franklie001]

Yes i 've remembered everything this time

 

[Franklie001]

Seriously thanks for the big help on this one

 

[Steve4Physics]

I've got Friction force = 1040N
and coefficient of kinetic friction u = 0.048

Well done. But you may have a rounding error. I got the friction force to be 1027.45N which rounds to 1030N.