# Homework Help: Forces on a body in contact with a horizontal plane

1. Oct 26, 2008

### eatingblaa

1. The drawing shows a 27.4-kg crate that is initially at rest. Note that the view is one looking down on the top of the crate. Two forces, and , are applied to the crate, and it begins to move. The coefficient of kinetic friction between the crate and the floor is k = 0.313. Determine the (a) magnitude and (b) direction (relative to the x axis) of the acceleration of the crate.

2. f=ma

Fk=UkFn

Fn=ma

3. I found the overall force to be 126.9 N 34.6 degrees to the x axis (the angle is verified to be correct)

I then found the kinetic friction to be 8.41N

Because they do not specify the static coffecient i think we leave that out.

total force - kinetic friction

126.9 - 8.41 =118.49

using f=ma

I found acceleration to be 4.32 ms-2

this is wrong I was wondeirng if anyone could tell me what i did wrong

2. Oct 27, 2008

### alphysicist

Hi eatingblaa,

I don't believe this is the correct force of kinetic friction. How did you get this number, and do you see what it needs to be?

3. Oct 27, 2008

### eatingblaa

I first found the normal force = mass x gravity = 268.794 N

Then kinetic friction = normal force x kinetic coeffecient = 8.41 N

is this right?

I dont know what you mean by do i see what it needs to be, could you please explain,

thank you

4. Oct 27, 2008

### heth

eatingblaa, imagine you were teaching physics to someone who hasn't done as much physics as you have. How would you explain what a "normal force" is to them?

Exploring this will help you with this problem (and many many others).

5. Oct 27, 2008

### eatingblaa

I would explain that because of newtons 3rd law of motion "to every action there is an equal and opposite reaction" The action of gravity creating the weight of the object must have an equal and opposite force, this is called the "normal force"

Would this be an okay explanation?

6. Oct 27, 2008

### alphysicist

Yes, this is correct.

This is correct up to here.

But this is not right. I was not sure what you had done, but now it looks like its just a calculation error.

The coefficient is around 1/3, so the frictional force should be around 1/3 of the normal force. Since the normal force is around 270N, the frictional force should be around 90N, so 8.41N is rather far off.

Last edited: Oct 27, 2008
7. Oct 27, 2008

### heth

Hmm. Probably best to put some time in exploring further, as it'll help your overall understanding. Your textbook should help you answer the first two questions - sketch and explore and see what you come up with for the others.

a. Do Newton's 3rd law force pairs act on the same object, or on different objects?

b. If you have a Newton's 3rd law force pair, can they be of different types, or do they have to be of the same type? (By 'type' I mean a contact force, a non-contact force, etc)

c. Imagine an unsupported crate falling through the air. What force is acting on the crate, and what type of force is it; where does this force come from?

d. What is the other force in the Newton's 3rd Law pair that "goes with" the force you described in part c?

e. (unrelated to what came before) Why is a normal force called a normal force? What does the word "normal" mean in this context?

8. Oct 27, 2008

### heth

The force F1 is helping to support the crate. So the floor isn't having to exert as large a supporting force as it would if F1 wasn't there. The normal force (which is the support force in this case) must be less than mg.

Edit: My apologies - I missed the 'top view' caption on the original diagram, so you can ignore this comment.

Last edited: Oct 27, 2008
9. Oct 27, 2008

### alphysicist

How is it helping to support the crate? F1 and F2 are horizontal forces.

10. Oct 27, 2008

### heth

Sorry. You're completely correct, I missed the "top view" caption on the diagram. I'll do an edit so as not to confuse the original poster.

11. Oct 27, 2008

### alphysicist

Oh, I see. I do hope eatingblaa still answers your questions, as it seems that they have a bit of misunderstanding of the third law.

12. Oct 27, 2008

### heth

13. Oct 28, 2008

### eatingblaa

your right it was a calculation error, I had taken the friction coefficient as 0.0313 rather than 0.313 creating a 1 decimal place error. Got it right now tho.

Thanks for your help and this has shown my lack of understanding, i just kinda know what to do, not why Im doing it if you understand.

Im going to try answer your questions before the day is over, to try and fully understand this. A bit busy tho,

thanks for the help again