When the truck is at rest, will the crate also be at rest?

[javii]

Homework Statement

I have attached the known information.

When the truck is at rest, will the crate also be at rest or will it slide
downwards?

Homework Equations

Sum F_y =0
G_max = (my_s)*N
G_min = (my_k) * N

The Attempt at a Solution

I did a Free body diagram on the crate (see the attached file)

The first step is finding the normal force.
Sum F_y =m*a_y =0
N-mg*cos(10)=1932.19 N

now i find the friction forces
G_ max=(my_s)*N= 0.70*1932.19=1352.53 N
G_min=(my_k)*N=0.50*1932.19=966.10 N

Now i will find F_II
F_II = m*g*sin(10) = 340.70 N

since G_max > F_II there is no motion downwards i,e a_x = 0

I am not sure if I have done it correct.
Thank you for your help.

 

[haruspex]

N-mg*cos(10)=1932.19 N

I assume you mean N=mg*cos(10)=1932.19 N

G_min = (my_k) * N

Only if it is moving. And it is mu, not my.
You can get the actual Greek character and subscript using the buttons just above the text area, Σ, X₂, X²

since G_max > F_II there is no motion downwards

Yes.

 

[javii]

I assume you mean N=mg*cos(10)=1932.19 N

Only if it is moving. And it is mu, not my.
You can get the actual Greek character and subscript using the buttons just above the text area, Σ, X₂, X²

Yes.

Yes i ment N=mg*cos(10)=1932.19 N (isolating for N)

So I don't have to calculate for G_min? Meaning I can skip this part in the calculations?

 

[haruspex]

Yes i ment N=mg*cos(10)=1932.19 N (isolating for N)

So I don't have to calculate for G_min? Meaning I can skip this part in the calculations?

Yes. You can assume the crate was placed statically. Until static friction is overcome, it is going nowhere.
I would have guessed there are more parts to the problem.

 

[javii]

Yes. You can assume the crate was placed statically. Until static friction is overcome, it is going nowhere.
I would have guessed there are more parts to the problem.

Yes. You can assume the crate was placed statically. Until static friction is overcome, it is going nowhere.
I would have guessed there are more parts to the problem.

Okay. Thank you very much. And yes there are more parts to the problem, i am trying to solve them now.