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What is the coefficient of static friction?

  • Thread starter Chica1975
  • Start date
  • #1
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Homework Statement


in a demonstration designed to determine the coefficient of friction for a book on a wooden plank, one end of the plank was raised until the book on it began to slide. if sliding commenced when the plank was at an agle of 37 degrees above the horizontal, what was the coefficient of static friction?


Homework Equations


Friction = constant * F
Fmg = Mass * 9.8



The Attempt at a Solution


where in the world do I start with his. I have no mass so I can't find mg. I have nothing to start from but the 37 degrees. I can't even try trigonometry becoz I have no other values. I know that the friction constant changes depending on the surfaces and this is a wooden one - am I supposed to memorize all the constants?

I have no idea

Can anybody shed some light on this - I have an exam tomorrow.
 

Answers and Replies

  • #2
PhanthomJay
Science Advisor
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Draw a free body diagram of the book and identify the forces acting on it. What forces act on it ? (there are 3). Then apply Newton's laws in the directions perpendicular and parallel to the plane. At 37 degrees, the book is just on the verge of sliding, but it is still in equlibrium in both those directions. You may not need to know the mass.
 
  • #3
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to be honest I have no idea i have drawn the free body diagram 3 times now and have tried to find numbers for the forces acting on it. First of newton's laws is inertia, no 2 is sumF= ma, third is that there is always a reaction to any action.

I can't see how I can derive the friction from any of the above?

There are these forces acting against this book-
mg
normal force
opposite to normal force
the force of the book moving down the plank
and the friction in the opposite direction

so tell me how on earth do you get the flippin friction - i am at my wits end
 
  • #4
Draw the free body diagram. friction = const * force acting normal to the surface.
here the force acting normal to the surface is mgcos(37). this will give friction force which will balance the component of mg which is parallel to the surface ie. mg sin(37)
Now let the constant be k
then,
kmgcos37 = mgsin37
k = tan37

that is how we get coefficient using angle
 
  • #5
63
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thank you, I understand everything except for the following

what do you mean by balance, and how does the parallel force balance kmgcos37?
 
  • #6
balance means - cancel the effect. If you apply a force on a block it will move. But if someone applies equal force in opposite direction on the same block then net force will be zero and the block will not move. tis is called balance
 
  • #7
63
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thank you very much - I got it out!
 

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