I am studying the solution used for the following question:
And I am unable to understand why
normal force Fn = mgcosΘ + PsinΘ
Should it not be just Fn = mgcosΘ, When drawing the FBD, the only forces that act in the Y direction are Fn and Fgcos(theta) , so Fn = -Fgcos(theta)
The Attempt at a Solution
Please see above.