Normal Force and Acceleration down the slope

AI Thread Summary
The discussion focuses on understanding the normal force and acceleration of an object on an inclined plane. The correct expression for the normal force is derived from the angle of the incline, with the normal force being perpendicular to the surface. The acceleration down the slope is determined by the component of gravitational force acting parallel to the incline, which is expressed as a = g sin θ. The confusion arises from the use of different angles; the derived angle must be used to accurately calculate forces. Drawing a force triangle helps clarify the relationships between the forces acting on the object.
lazyguy91
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Homework Statement


normalforce.jpg

2. Homework 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
 
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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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