Static friction on an inclined plane.

AI Thread Summary
The discussion focuses on the calculation of static friction on an inclined plane. The original confusion arises from the distinction between the frictional force and the normal force, with the correct frictional force being f = mgsin(theta) rather than f = mgcos(theta). Participants clarify that the component of weight acting along the incline is responsible for the block's motion down the slope. The frictional force opposes this motion, highlighting the importance of understanding the forces at play on an inclined plane. Accurate identification of these forces is crucial for solving problems related to static friction in physics.
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Homework Statement



http://img124.imageshack.us/img124/9997/problem3.jpg



Homework Equations



f=mgcos theta


The Attempt at a Solution


I am thinking the answer is f = mgcos theta because gravity only acts on y-component but the answer is f=mgsin theta . Can anyone explain?
 
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In the problem, f is the frictional force ,not the normal reaction.
Frictional force acts in the opposite direction of the motion of the block.The block moves down due to the component of weight along the inclined plane. What is that component?
 
rl.bhat said:
In the problem, f is the frictional force ,not the normal reaction.
Frictional force acts in the opposite direction of the motion of the block.The block moves down due to the component of weight along the inclined plane. What is that component?

x-component of weight?
 
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