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

In summary, the truck will not move when the crate is at rest because there is a static force of 1932.19 Newtons acting against it.
  • #1
javii
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0

Homework Statement


I have attached the known information.
problem-2-png.113891.png

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.
 

Attachments

  • problem 2.PNG
    problem 2.PNG
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  • #2
javii said:
N-mg*cos(10)=1932.19 N
I assume you mean N=mg*cos(10)=1932.19 N
javii said:
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
javii said:
since G_max > F_II there is no motion downwards
Yes.
 
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  • #3
haruspex said:
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?
 
  • #4
javii said:
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.
 
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  • #5
haruspex said:
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.
haruspex said:
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.
 

1. What is the definition of "rest" in this scenario?

In this context, "rest" refers to the state of an object when it is not moving or experiencing any acceleration.

2. Will the crate move if the truck is stopped but on an incline?

Yes, the crate will move due to the force of gravity acting on it. The truck being at rest does not necessarily mean that the crate will also be at rest.

3. What if the truck is at rest but the crate is on wheels?

The crate will not be at rest in this scenario, as it is still capable of moving due to the wheels. The truck being at rest does not affect the motion of the crate in this case.

4. Can the crate be at rest if the truck is moving at a constant velocity?

Yes, the crate can be at rest if the truck is moving at a constant velocity. This is because the crate is also moving at the same velocity as the truck, so it is not experiencing any acceleration.

5. Is there any other factors that could affect whether the crate is at rest when the truck is at rest?

Yes, other factors such as external forces (e.g. wind, friction) or the presence of another object (e.g. a person pushing the crate) could affect whether the crate is at rest when the truck is at rest. These factors can introduce additional forces that may cause the crate to move or remain at rest.

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