Finding Magnitude of Unkown Force on Rough Horizontal Surface

  • Thread starter FaraDazed
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  • #1
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Homework Statement


A block of mass 100g is at rest on a rough horizontal plane. It has a massless rope attached to one end inclined at 20° above the horizontal. When the tension in the string is 0.5N the object is found to be in limiting equilibrium.

Part A: Find the coeffecient of static friction between the object and the plane.

Part B: What would the tension in the string have to be to make the object accelerate at 1.5m/s^2.


Homework Equations


Fnet=ma
R=mg (R is the normal reaction force)


The Attempt at a Solution


For clarification, F_f is the frictional force. and P is the unkown foce in part B.

Part A:
[itex]
0.5cos20=0.47\\
0.5sin20=0.17\\
mg=R+0.17\\
0.98=R+0.17\\
∴ R=0.98-0.17=0.81\\

F_f=0.5cos20 (because \space the \space force \space applied\space by\space the\space string\space to\space the\space horizontal \space must \space equal \space the \space frictional \space force)\\
∴F_f=0.47\\
μR=0.47\\
μ0.81=0.47\\
∴ μ \frac{0.47}{0.81}=0.58 \\
[/itex]

Part B:
[itex]
R+Psin20=mg \\
R+Psin20=0.98 \\
∴ R=0.98-0.34P \\
\\
Pcos20-F_f=ma\\
Pcos20-μR=0.15\\
∴0.94P=0.15+0.58R\\
0.94P=0.15+0.58(0.98-0.34P)\\
0.94P=0.15+0.57-0.2P\\
1.14P=0.15+0.57\\
1.14P=0.72\\
∴ P=\frac{0.72}{1.14}=0.63N \\
[/itex]

My main issue is if I have done it right because the answer to Part B seems rather low compared to the force when it is in limiting equilibrium i.e. only 0.13N more.

Any help/pointers is appreciated.
 
Last edited:

Answers and Replies

  • #2
haruspex
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Hi Faradazed,
As a matter of technique, it's a really good idea to do most of the work purely in algebra, using symbols even for values that are known. Then plug all the numbers in at the end. It's much easier to spot mistakes, and less prone to silly errors.
Ff =0.81
I assume you meant Ff = .5 cos 20
Part B:
The question is not quite kosher here. It should tell you the dynamic friction.
the answer to Part B seems rather low compared to the force when it is in limiting equilibrium i.e. only 0.13N more.
You know it will be roughly .15N more, since that would be the force to accelerate it that fast with no gravity or friction. You may be surprised that it's a bit less than .15. That's because some of the extra force goes into reducing the normal force, and hence the friction. That 'frees up' some of the 0.5N to produce acceleration. In general, there is an optimal angle at which to drag a load.
 
  • #3
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Hi, thanks for replying :)

Hi Faradazed,
As a matter of technique, it's a really good idea to do most of the work purely in algebra, using symbols even for values that are known. Then plug all the numbers in at the end. It's much easier to spot mistakes, and less prone to silly errors.
Yes it is a habit I need to get out of, my math teacher has commented on it, I am trying to improve upon this :)

I assume you meant Ff = .5 cos 20
Yes I think this was a copying error from my written work to typing it onto the computer
The question is not quite kosher here. It should tell you the dynamic friction.
Is that the same as kinetic friction? We have only been taught about static friction so I assume all problems given are meant to act like as if kinetic friction is nonexistent.
You know it will be roughly .15N more, since that would be the force to accelerate it that fast with no gravity or friction. You may be surprised that it's a bit less than .15. That's because some of the extra force goes into reducing the normal force, and hence the friction. That 'frees up' some of the 0.5N to produce acceleration. In general, there is an optimal angle at which to drag a load.

I see. Does what I have done (apart from not writing it out in an algebraic way to start with) look OK to you then?
 
  • #4
haruspex
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Is that the same as kinetic friction?
Yes, kinetic is the usual word, sorry.
We have only been taught about static friction so I assume all problems given are meant to act like as if kinetic friction is nonexistent.
I would think the default assumption is that it's the same as the static friction. That seems to be what you have to assume here.
Does what I have done (apart from not writing it out in an algebraic way to start with) look OK to you then?
Yes, it's fine.
 
  • #5
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Yes, kinetic is the usual word, sorry.

I would think the default assumption is that it's the same as the static friction. That seems to be what you have to assume here.

Yes, it's fine.

OK, many thanks :)
 

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