Static friction coefficient for a crate in the back of a truck

[purple4]

Homework Statement

John puts a crate in the back of his truck. The maximum acceleration of the truck is 4.2m/s^2. What is the minimum coefficient of static friction so that the crate doesn't slide?

Relevant Equations

F=ma
Fg=mg
μk=Fk/N

I drew a free body diagram for the crate. I have normal force going up, weight down, static friction force to the left and the force of the crate sliding to the right. I'm assuming the truck is moving to the left.

I have Fnet=0 because the crate is not moving, so the forces are balanced. I think that the force of the crate sliding backward would be equal to the force of the truck moving forward, but I don't know how to get a force in Newtons from the acceleration value.

I know that I need to find the normal force as well as the static friction force in order to find the static friction coefficient, but I'm not sure how to do that since Fnet=0 and I don't know the mass of the crate.

Any help would be much appreciated.

 

[BvU]

Hello @purple4 ,

$\qquad$!​

I drew a free body diagram for the crate.

Can we see it ?

the force of the crate sliding to the right.

I thought it wasn't supposed to slide ?

Fnet=0 because the crate is not moving

You mean: is not moving with respect to the truck, I hope ?

So with respect to the earth, what's the situation ?

I don't know the mass of the crate

Call it $m$ and cross your fingers it divides out ... $\$

 

[purple4]

Thanks for replying! Above is my free body diagram.

It’s not sliding, but I put the force Fx opposite to the static friction because I thought the forces have to be balanced since there’s no motion of the crate with respect to the truck.

It’s not moving with respect to the truck, but with respect to the Earth it’s moving the same direction as the truck. I'm just confused about how the acceleration of the truck goes with the information about the crate.

And by that last part do you mean to put m in a formula twice so it cancels out?

 

[haruspex]

I put the force Fx opposite to the static friction because I thought the forces have to be balanced since there’s no motion of the crate with respect to the truck.

You can't just go inventing forces like that.

You need to pick a reference frame.

1. If you choose the ground then the crate is accelerating, so the horizontal forces are not balanced. There is no need to invent Fx.

2. If you choose the truck as your reference frame then that is accelerating, so non-inertial. When you choose a non-inertial frame you need to add 'fictitious' forces, such as centrifugal and Coriolis forces. In the case of a linearly accelerating frame, as here, you add a force that would result in an equal and opposite acceleration. So if the truck has acceleration $a$ to the left and the crate has mass $m$ then you add an inertial force $ma$ to the right. Since the crate is not accelerating in the reference frame of the truck, that will balance the frictional force, i.e. it is the Fx in your diagram.

do you mean to put m in a formula twice so it cancels out?

Not quite. @BvU is saying that if you just write out the equations, putting the crate's mass as m, then there is an excellent chance that m will occur twice in such a way that it cancels.

 

[purple4]

Ok, I think I’ve got it. I used μs=Fs/N, but I think Fs=Fnet so I replaced Fs with ma and N with mg. Then m cancels out and I end up with μs=a/g.

 

[BvU]

So now your FBD can do without $F_x$.
The vertical forces $mg$ and $N$ cancel each other, so no net vertical force. And $F_s = \mu_s N = \displaystyle {a\over g} m g = ma \$, a net force $ma$ to the left and the crate accelerates with the same acceleration $a$ as the truck, and in the same direction !

$\$

 

[purple4]

Ok, got it! Thank you!