Is the upward Ff of a stationary block on an incline more or less than mg?

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The discussion revolves around the forces acting on a stationary block on an incline. It clarifies that the force acting up the ramp is less than the gravitational force (mg) due to the component of gravity acting down the slope, which is mg sin(x). The block remains at rest because the static friction force balances this component of gravity, preventing it from sliding down. The normal force also contributes to the overall force dynamics but does not exceed mg. Understanding these forces is crucial for grasping the equilibrium of objects on inclined planes.
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A block of mass m remains at rest on an incline. The force acting up the ramp on this block is
a. 0
b. mg
c. less than mg
d. more than mg

I understand that the force of friction cannot be more than the force of gravity because it would then move upward... but how is that the force of friction is less than (mg)? Wouldn't the block then slide down?

Of course it wouldn't because of the mgsin(x)... but I just need some clarification more than anything. Trouble to grasp for some reason.
 
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Hello mirandab17 . Welcome to PF !

A component of the normal force is vertical, so ...
 

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