Register to reply

Mg cos theta in context of incline plane

by negation
Tags: context, incline, plane, theta
Share this thread:
negation
#1
Mar31-14, 01:04 PM
P: 819
For a mass on an incline plane, is the downward force mg cos Θ or -mg cos Θ?
I'm inclined to state there should be a negative but from my lectures, the negative does not appear to be stipulated. I was wondering if it's was a mistake by the lecturer.

mg cos Θ + FN = 0
mg cos Θ = -FN

the downward force must be of equal magnitude to the normal force-but opposite direction.
This makes sense mathematically.

However, geometrically, it does make sense for the down ward force to be -mg cos Θ.

Could someone clarify?
Phys.Org News Partner Physics news on Phys.org
Step lightly: All-optical transistor triggered by single photon promises advances in quantum applications
The unifying framework of symmetry reveals properties of a broad range of physical systems
What time is it in the universe?
ModusPwnd
#2
Mar31-14, 01:11 PM
P: 1,067
It depends! Are you defining up or down to be negative? You are free to do either. But the great thing about vectors is that it doesn't matter what coordinate frame you choose. And you have just found this out. Regardless of your choice of positive or negative you have found that the normal force must be in an opposite direction of the gravitational force. Im not sure if that clarify's completely but it should help.
Doc Al
#3
Mar31-14, 01:11 PM
Mentor
Doc Al's Avatar
P: 41,471
Using a sign convention where into the plane is negative, the component of the weight normal to the surface would be -mg cosθ:
-mg cosθ + FN = 0
FN = mg cosθ (positive, thus out of the plane)

negation
#4
Mar31-14, 01:21 PM
P: 819
Mg cos theta in context of incline plane

Quote Quote by ModusPwnd View Post
It depends! Are you defining up or down to be negative? You are free to do either. But the great thing about vectors is that it doesn't matter what coordinate frame you choose. And you have just found this out. Regardless of your choice of positive or negative you have found that the normal force must be in an opposite direction of the gravitational force. Im not sure if that clarify's completely but it should help.
Quote Quote by Doc Al View Post
Using a sign convention where into the plane is negative, the component of the weight normal to the surface would be -mg cosθ:
-mg cosθ + FN = 0
FN = mg cosθ (positive, thus out of the plane)
I am defining down to be negative.

I am suspecting the lecturer committed a blunder by forgetting the negative sign.


Register to reply

Related Discussions
Why does a=g*sin(theta) on an inclined plane Classical Physics 6
How to prove Theta for a incline plane. General Physics 2
Finding theta on an incline plane Introductory Physics Homework 1
Frictionless plane at angle of theta=69degrees Introductory Physics Homework 1
Non slip ball down an incline of theta Introductory Physics Homework 3