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fishpaste
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#1
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#1
You have a plank of wood resting on a smooth barrel and touches a rough floor. The plank weighs 3kg, and is 2m.

Find the force exerted on the plank by the barrel?
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fishpaste
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#2
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#2
Coefficient of friction between plank and floor is 0.8
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way2go
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#3
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#3
What is the height of the barrel? I think you need it.
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JaF
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#4
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#4
0.8*3*g

Code:
----------> reaction from barrel
| \
|  \
|   \
|    \
|     \
|      \
|    |  \
|    |   \
|    |    \    ^
|    |     \   |
|    V      \  | Reaction
|  weight  \ |
    <------
     Friction
By resolving horizontally friction = reaction from barrel

since friction = 0.8*R

and R = 3g

reaction from barrel = 0.8*3*g
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fishpaste
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#5
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(Original post by way2go)
What is the height of the barrel? I think you need it.
Not given :/ I suspect it's done with moments.
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fishpaste
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#6
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#6
(Original post by JaF)
0.8*3*g
No, it rests against the floor and the barrel.

I've just realised I did miss a crucial detail, I'm so sorry, the angle between the barrel and the point it's resting at is 60 degrees, however the distance from the far end of the plank and the point of contact is not given, and so I don't think you can work out the height.
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JaF
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#7
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#7
Is this what it looks like?



Edit: the image has been chopped off I'll change it
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way2go
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#8
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#8
(Original post by JaF)
Is this what it looks like?



Edit: the image has been chopped off I'll change it
Hey, I was doing exactly the same thing. I am not sure what the situation is like. Here is a sketch I made. Does it agree with the problem?

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fishpaste
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#9
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#9
I think the reaction force of the barrel acts at an angle :/
Code:
\
      \
   __  \
 /     \\
|      | \
 \___/    \
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JaF
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#10
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#10
That picture isn't working mate
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way2go
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#11
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(Original post by JaF)
That picture isn't working mate
Yes I see. I am working on it.
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Ben.S.
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#12
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(Original post by JaF)
That picture isn't working mate
The NORMAL reaction, exerted by the barrel on the plank is NORMAL to the plank at the point of contact. The only other things to take into account are the force of friction (acting on the plank where it touches the ground) and the weight of the plank. Then you can simply resolve forces and, given an angle, find the reaction.

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way2go
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#13
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#13
here is the picture:

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way2go
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#14
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Fishpaste, is this picture correct? If it is we can start calculating...
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Ben.S.
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#15
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0.75N, I get.

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fishpaste
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#16
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#16
The picture is not correct, the plank goes past the barrel, and the coefficient of friction is 0.8, not the friction force.
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Ben.S.
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#17
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(Original post by fishpaste)
The picture is not correct, the plank goes past the barrel, and the coefficient of friction is 0.8, not the friction force.
Ergh - giving wrong pictures really doesn't help, you know. I thought it was a bit easy - what with not having to use the coefficient of friction and all.

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Ben.S.
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#18
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(Original post by Ben.S.)
Ergh - giving wrong pictures really doesn't help, you know. I thought it was a bit easy - what with not having to use the coefficient of friction and all.

Ben
Not your fault, though. I used someone else's picture with your numbers.

Ben
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fishpaste
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#19
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#19
(Original post by Ben.S.)
Ergh - giving wrong pictures really doesn't help, you know. I thought it was a bit easy - what with not having to use the coefficient of friction and all.

Ben
Sorry, tried to show it going past on my picture.
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way2go
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#20
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#20
(Original post by fishpaste)
The picture is not correct, the plank goes past the barrel, and the coefficient of friction is 0.8, not the friction force.
Yes, I know. The friction force is 0.8 x normal force (0.8*N) and not 0.8 Newton. I am sorry it was a bit confusing.

I hope you can work out the problem yourself, because I really cannot solve it without a decent description or sketch of the problem. Good luck with it.
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