Is the Frictional Force on a Sloped Block Greater Than Its Weight?

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The discussion revolves around determining the relationship between the frictional force and the weight of a block resting on a rough incline. Initially, there is confusion about whether the frictional force is greater than, equal to, or less than the block's weight. It is clarified that the frictional force should be equal to the component of the weight acting along the slope, not greater. A misunderstanding about the decomposition of gravitational forces is acknowledged, leading to the realization that the initial calculations were incorrect. Ultimately, the correct approach involves accurately analyzing the forces acting on the block.
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Homework Statement


A block is at rest on a rough incline .
The frictional force acting on the block, along the incline, is
A) greater than the weight of the block.
B) equal to the weight of the block.
C) zero.
D) less than the weight of the block.

Homework Equations

The Attempt at a Solution



The frictional force should be equal to the Weight along the slope so MG/sin(theta) so shouldn't the frictional force be greater than the weight of the block?
 
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The weight along the slope is greater than the total weight?

How do you get MG/sin(theta)? That would imply, as the angle theta goes to zero, the weight along the slope goes to infinity. Is that reasonable?
 
I know I'm doing something wrong but i don't know what
 

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Nvm I got it I was decomposing my gravitational force vector incorrectly
 

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