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Normal Force and Acceleration down the slope 
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#1
Dec1909, 12:39 PM

P: 1

1. The problem statement, all variables and given/known data
2. Relevant equations + questions 1) Normal Force when considering mg as the vertical force, mg = N cos θ however, when considering N as the vertical force, N = mg cos θ note: θ is the same due to opposite angle which equation is correct? 2) Acceleration down the slope due to gravity (frictionless) a = g sin θ why can't I use the originally given θ? e.g. sin θ = g/a a = g/sin θ why must I use the derived θ instead of the one between the horizontal and the inclined place? where are my mistakes? Please reply as soon as possible. Thanks 


#2
Dec1909, 12:54 PM

P: 1,089

I don't know what your trying to do with all those equations and such, just try to understand the normal force.
The normal force is always going to be perpendicular to the plane on which the object lies. (Your drawing depicts this accurately) We know that since its on an inclined plane its not going to simply just be N = mg(flat plane), its going to be some variation of that because of the angle of the plane. So how do we figure this out? Well just simply draw a force triangle and solve it. You're going to have mg going straight down as the hypotenuse and the other two sides will be determined based on the angle of your plane. ( mgcos(theta) and mgsin(theta) ) [ You should find that mgcos(theta) is in the opposite direction of the normal force ] As for part 2), you can also answer this once you've drawn your force triangle, mgsin(theta) should be acting parallel to the plane (depending on how you've defined your x and y axis). 


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