Determine the magnitude of the total force exerted by the track on the

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SUMMARY

The discussion centers on determining the total force exerted by a track on a particle in motion, specifically addressing the confusion surrounding the direction of the normal force in a Free Body Diagram (FBD). The participant questions why the normal force is depicted downwards in the provided FBD, while their understanding suggests it should be upwards due to the gravitational force (mg). The solution clarifies that the direction of the normal force can vary based on speed and radius, and emphasizes that the sign of the normal force does not affect the magnitude of the frictional force.

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TheDudeTR
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Homework Statement
i did not understand free body diagram in this question
Relevant Equations
F = ma
This is the question.
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243433

This is the FBD in the solution given. What i didn't understand is the sense of normal force. in the question mg force pulls particle down and according to my logic normal vector should be upwards in this conditions. but it is downwards in the given FBD. when i try to solve question with my own way result is different because of in the question it takes normal axis equations as (N+mg.cos30=m*a) and i took it as (-N+mg.cos30=m*a ) where am i wrong? can you help me please.
 
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TheDudeTR said:
the sense of normal force
In principle it could be either way - depends on the speed and the radius, etc.
But you do not need to care which way it is when drawing the diagram. If you chose wrongly it will come out negative, but the frictional force only cares about the magnitude.
 

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