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

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SUMMARY

The discussion centers on the forces acting on a stationary block of mass m on an incline. The force acting up the ramp is less than the gravitational force mg due to the component of gravitational force acting parallel to the incline, specifically mg sin(x). The block remains at rest because the static friction force balances this component, preventing motion down the incline. Understanding the relationship between gravitational forces and friction is crucial for analyzing static equilibrium on inclined planes.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Basic knowledge of static friction and its role in equilibrium
  • Familiarity with trigonometric functions, particularly sine
  • Concept of normal force and its components on inclined planes
NEXT STEPS
  • Study the concept of static friction and its maximum value
  • Learn about inclined plane dynamics and the role of angles
  • Explore free body diagrams for objects on inclines
  • Investigate the effects of varying incline angles on static equilibrium
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Students of physics, educators teaching mechanics, and anyone interested in understanding forces on inclined planes and static equilibrium.

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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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