Confuse about mg cos@ = R or R cos@ = mg

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SUMMARY

The discussion centers on the equilibrium of forces acting on a body placed on an inclined plane at an angle θ. It is established that the normal reaction force (R) balances the component of gravitational force acting perpendicular to the surface, specifically R = mg cos(θ). The participants confirm that there is no acceleration perpendicular to the inclined surface, reinforcing the conclusion that R must equal mg cos(θ) for equilibrium to be maintained.

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Scharles
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i am quite confusing about those reaction force...
let say a body is placed on inclined plane with angle @...
at equilibrium, what are the forces that balances each other ??

mg cos@ balanced by R or R cos@ balance mg ??

thanks~
 
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Hi Scharles! :smile:

(have a theta: θ :wink:)

I always work it out each time …

you know there can be no acceleration perpendicular to the surface …

so the forces perpendicular to the surface must add to zero …

so R balances mgcosθ. :smile:
 
Hi Scharles! :smile:
Scharles said:
does my forces diagram draw wrongly ??

No, your diagram looks fine (and very cheerful! :biggrin:).

(btw, it's "is my forces diagram drawn wrongly", or "have I drawn my forces diagram wrongly" or "does my forces diagram look wrong" :wink:)
 

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