Find the direction and the magnitude of Frictional force

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Homework Help Overview

The discussion revolves around understanding the direction and magnitude of the frictional force acting on a block, particularly in the context of forces at play when the block is at rest. The subject area includes concepts of friction, reaction forces, and equilibrium in mechanics.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the role of the reaction force in relation to the frictional force, questioning why it may not be considered in certain calculations. There is a discussion on the relevance of limiting friction and the conditions under which different forces are resolved.

Discussion Status

The discussion is ongoing, with participants providing insights into the relationship between the forces acting on the block. Some guidance has been offered regarding the conditions of equilibrium and the importance of parallel forces, though there is no explicit consensus on the role of the reaction force.

Contextual Notes

There is mention of missing information, such as the value of the coefficient of friction, which may affect the analysis. Participants are also considering the implications of the block being at rest in their reasoning.

chwala
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Homework Statement
See attached Question 5.
Relevant Equations
Mechanics
1716084035512.png


In my working i have, the attached. My question is Why is the Reaction force not being considered here,

My equations are;

Frictional force = ## 30\cos 50^0 - 20\cos 40^0 = (19.28 - 15.32)N = 3.96N##
The direction will be downwards.

1716084093643.png
 

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The reaction force is perpendicular to the direction of freedom of movement of the block.
 
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Lnewqban said:
The reaction force is perpendicular to the direction of freedom of movement of the block.
Ok a bit confusing...then if we were to talk of Limiting friction that's when the Reaction force comes into play...
like in this example;

1716092015168.png




aaaargh i think okay! It's only the parallel forces being resolved in both scenarios.
 
Last edited:
The key phrase here is "the block... which is at rest".
Meaning all forces acting along the direction of freedom of movement cancel each other.
Not much to do with the normal forces, because the value of the coefficient of friction is not given.


Block pressured against incline.jpg
 
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