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Inclined plane and force of friction

  • Thread starter dranseth
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1. Homework Statement
I am having some difficulties finding out the force of friction for an inclined plane. If we know the mass of the object, the angle of the inclined plane, and the acceleration, how to we find Ff?

I know how to draw the diagrams, but I can't seem to find what force of friction is without a coefficient of friction or an applied force.


2. Homework Equations



3. The Attempt at a Solution
 

Answers and Replies

rock.freak667
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I am assuming that the object is moving down the plane so that acceleration you have times the mass gives the resultant force down the plane.
The component of the objects weight acts down the plane...and the frictional force opposes this motion(i.e. acts up the plane). Can you now form an equation for the resultant force from these two forces?
 
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That makes sense. How do we know how many components to use for net force in a given problem? I was making this much more complicated than it is..

This is partially because I am teaching this to myself and I came across one question whcihO involved the sum of the applied force, parallel component to the gravitational force, and applied force. I didn't really understand why... It was in equilibrium so the added to equal 0.
 
rock.freak667
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That makes sense. How do we know how many components to use for net force in a given problem?
When dealing with problems involving inclined planes, spit the weight of the object into its components (One will always be perpendicular to the plane and the other will always be parallel to it acting down the plane).

After you have done that, consider the direction of motion. If the object does not move then the resultant force acting on it is zero. If the object moves down the plane then the resultant force is the sum of the forces acting down the plane(in the direction of motion)- the sum of the forces opposing motion(i.e. up the plane)

This is partially because I am teaching this to myself and I came across one question whcihO involved the sum of the applied force, parallel component to the gravitational force, and applied force. I didn't really understand why... It was in equilibrium so the added to equal 0
Post the question and I will see if I can help you understand it better
 
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You slide a 325N trunk up a 20degree inclined plane with constant motion by exerting a force of 211N parallel to the inclined plane.

what is the sum of the applied force, friction, and the parallel component to the trunks weight? why?


thanks in advanced
 
rock.freak667
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You slide a 325N trunk up a 20degree inclined plane with constant motion by exerting a force of 211N parallel to the inclined plane.

what is the sum of the applied force, friction, and the parallel component to the trunks weight?
ok well the resultant force is ma

the component of the 325N against motion is 325sin20

so that ma=211-325sin20-F (F is frictional force)...I hope you reached that far

The question stated that motion was constant. Remember what Newton's first law says about if a body moves with constant motion in a straight line?
 
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net=0
 
rock.freak667
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Right. so that the net force is zero

so that the sum of the forces is zero. But they did say that there is motion (which is constant). Now if the net force is zero,what does that mean for the acceleration? And hence velocity?
 
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a=0, velocity=constant
 
rock.freak667
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one last question. If I was dealing with a system that contained three blocks attached by a string such as:

[]-[]-[]

and they were accelerating in the right direction, when I am calculating the total frictional force, do I add up all of the normals for the equation Ff=ufn?
 
rock.freak667
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Yes you would.
 

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