Force required to keep a wooden block from sliding on a rough incline

  • Thread starter nishantve1
  • Start date
  • #1
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Homework Statement



This thing was a part of an entire question . I solved the other parts but I don't know why I am messing up here .

Force required to keep a wooden block from sliding on a rough incline with coefficient of friction μ

Homework Equations



Force on a block on an inclined plane ,
towards the ground : m*g*sin(θ)
due to friction in opposite direction of the above force is m*g*cos(θ)
where m is the mass of the block , g is the gravitational acceleration and θ is the angle the incline makes with the horizontal.

The Attempt at a Solution




So I pictured something like this
Initially
http://awwapp.com/s/38/18/09.png [Broken]

Now if I apply a force to stop the block from sliding the FBD should be something like this
http://awwapp.com/s/5a/44/08.png [Broken]

So the force I have to apply ( the blue one ) should be Force due to gravity - Force due to friction
F = mg(sin(θ) -μcos(θ))
But the answer given is
F = μ*m*g*cos(θ)
How is this possible
Sorry about the drawings
 
Last edited by a moderator:

Answers and Replies

  • #2
Doc Al
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I'd say you were correct. If this is a textbook problem, what textbook is it?
 
  • #3
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@Doc Al It's actually a Test prep book published by a very well known publisher in my country at least . This question was part of the actual test held in 2012 . The question goes like
The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is 1/(2*sqrt(3)) angle of the inclined plane is
a> 60 degrees
b> 45 degrees
c> 30 degrees
d> 15 degrees

The answer in the key is 30 degrees
 
  • #4
Doc Al
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The answer in the key is 30 degrees
I agree with that answer.
 

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