Find the direction and the magnitude of Frictional force

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SUMMARY

The discussion focuses on calculating the frictional force acting on a block, specifically using the equation Frictional force = 30cos(50°) - 20cos(40°), resulting in a magnitude of 3.96N directed downwards. The conversation highlights the role of the reaction force, which is perpendicular to the block's movement direction and is not considered in this scenario since the block is at rest. The importance of limiting friction is also noted, indicating that the reaction force becomes relevant when analyzing forces in motion.

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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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