What is the acceleration of the box?

  • Thread starter nesan
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  • #1
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Help Please. :)

Homework Statement


A 20 KG box is dragged across a level floor with a force of 100 N. The force is applied at an angle of 40 degrees above the horizontal. If the coefficient of kinetic friction is .32, what is the acceleration of the box?

Homework Equations


...Sorry, nothing.


The Attempt at a Solution


I tried and failed few times. Our teacher did give us the answer which is 1.8 m/s squared.

I need help getting to that answer. Thank you. :smile:
 
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Answers and Replies

  • #2
658
2


What are all the forces on the box? Draw a free body diagram and list all the x forces and all the y forces.
 
  • #3
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[PLAIN]http://img696.imageshack.us/img696/6268/98211780.png [Broken]

That's as far as I can go. I'm having difficulties with the angle. Help please. :frown:
 
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  • #4
658
2


Haha, PAINT! I'll use Paint too, I never thought of that.

Stratch what you did so far. Just worry about the forces not the numbers yet.

Can you see the other forces now?
 

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  • #5
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There's

- Friction
- Normal Force
- Force of gravity
- 100 N Drag Force

That's all. I though force of gravity gets eliminated by normal force. :confused:

Did I name them all?

:confused:
 
  • #6
658
2


Yes you named them all. Now N = mg if there isn't any other Forces in the y direction, right? If you break down the pull Force in this case it has 2 components, one in the x and one in the y. So, for this problem N ≠ mg. Here's a picture of the pull Force broke down into it's components. It's not moving in the y direction so all those forces equal 0.

So the question is: N = mg + ?
 

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  • #7
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Okay

So N = mg + F (the hypotenuse)

Shouldn't it be N = mg - F?

Since by pulling, a small force from mg is being cancelled out.
:confused:
 
  • #8
658
2


That was, er, a trick question...:blushing:

Ok, you're very close, it wouldn't be the hypotenuse it's only the component of F in the y direction:

N = mg - Fy

How can we write Fy?
 
  • #9
75
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We know the angle is 40 degrees. I'm sure there's some trigonometry involved here.:uhh:

I have no idea. Sorry. Can you tell me please? :shy:

Or we use Fnet = M * A

Fnet = Sum of all forces


??

:shy:
 
  • #10
658
2


Yeah, problems like this involve Trig a lot. Remember: SOH CAH TOA, it's a life saver.

In this case, since it's the side opposite of the angle you'd use Sin.
 
  • #11
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So it would be like

Sin 40 = O / H ----> Sin 40 * H = O

I'm sorry, can you show me how to do it. :frown:
 
  • #12
658
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Sin θ= O/H
H*Sin θ = O

So it's H*Sin 40 = O
 
  • #13
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Can you explain to me what comes next? I'm really confused. Sorry. :(
 
  • #14
658
2


All in all you need to find all the Forces in the x direction and all the Forces in the y direction. We've got all the Forces in the y direction done. Now for the x direction. Since the object is moving in the x direction these Forces *don't* sum to zero.
 
  • #15
75
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So friction is Mu * N

Since N = mg - Fy

Fnet in the x direction = 100 - Mu * N

Which = 100 - Mu * ((9.8 * 20) - Fy)

Now, I'm stuck. :(
 
  • #16
658
2


Refer back to the F breakdown into components the F in the x direction wouldn't be 100 it would be the Fx one. So that means Trig once again. In this case it's adjacent to the angle so we use Cos = A/H then solve for A so we have:

H*Cos 40 = A
 
  • #17
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Okay, I get it. :)

Since we don't know the hypotenuse how will we solve it?
 
  • #18
658
2


We do know the hypotenuse, it's 100N
 
  • #19
MysticDude
Gold Member
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I was able to do it in a different way without involving trigonometry and I get a = 1.864 m/s². But let's see how DrummingAtom's way works out.
 
  • #20
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H*Cos 40 = A

Stupid me. :(

100*Cos 40 = 70.6 N

and

H*Sin 40 = 64.2

Now we have all the sides.

So...

N = mg - Fy

N = ((9.8 * 20) - Fy)
N = 196 - Fy

N = 196 - 64.2

N = 131.8

Friction = Mu * N

Friction = 0.32 * 131.8

Friction = 42.176

Fnet = Fx - Friction

Fnet = 70.6 - 42.176

Fnet = 28.424



Fnet = M * A

28.424 = 20 * A

28.424/20 = A

A = 1.4212

It's close to 1.8 m/s squared.

Is that okay? And did I do it right? :)
 
  • #21
MysticDude
Gold Member
140
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It is close but I wouldn't want to take it. I understand DrummingAtom's way of doing it but here is what I did:

F - f = ma. F is the force that is applied and f is the friction.
Well, we already know F,f, and m and we only have to find a. So with some algebra-fu we get: [tex]\frac{F-f}{m} = a[/tex]. We also know that the formula for friction is μkFN where FN is the normal force.
So we just substitute in the numbers and get: [tex]\frac{100 - .32(20*9.8)}{20} = a[/tex]. I get 1.864. Now I'm not completely sure on the normal force part so I'll be double checking my work again.

EDIT: I just did the way that DrummingAtom said to do and I get a = 1.722 which I guess is a reasonable answer too. Here is how I did it:
Well we already know that FN is 131.721239. Check!
So then we also know that Fx is 100cos(40°) which is 76.6044. Check! You had 70.6 and I think that this is the reason why you got 1.4212 instead of the desired 1.8.
Anyway...
We continue with the fact that Fnet = Fx - μkFN to get 34.4536 N.
After that we use the ol' a = F/m to get the answer. So just do 34.4536/20 to get 1.7227 m/s²!
 
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  • #22
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Awesome ;D

Thank you so much guys. :biggrin:
 
  • #23
MysticDude
Gold Member
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Awesome ;D

Thank you so much guys. :biggrin:

Well I hope that you understand the way that DrummingAtom did it because I think he did it the "more" correct way. I think I either got lucky OR just did it another way.
 
  • #24
75
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I understand it clearly and thank you very much you too. ^^
 

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