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

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In the discussion about forces on an inclined plane, it is clarified that at equilibrium, the normal force (R) balances the component of gravitational force acting perpendicular to the surface, which is mg cosθ. There is an emphasis on the necessity for no acceleration perpendicular to the surface, leading to the conclusion that R = mg cosθ. The participant confirms their forces diagram is correct, and the conversation includes light-hearted commentary on phrasing. Overall, the key takeaway is the relationship between normal force and gravitational force on an incline.
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:​
 
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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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