- #1
Solve FBD Problem: M.asinθ=N, N.sinθ=Ma
- Thread starter hav0c
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Homework Help Overview
The discussion revolves around a free body diagram (FBD) problem involving forces and accelerations, specifically focusing on the relationships between normal force, mass, and gravitational force in a system influenced by an angle θ. Participants are examining the correctness of various equations derived from the FBD.
Discussion Character
- Mixed
Approaches and Questions Raised
- Participants are exploring the relationships between forces and accelerations, questioning the validity of specific equations related to the normal force and gravitational force. There is a focus on the horizontal components of forces and accelerations, with some participants expressing uncertainty about the correctness of their derived equations.
Discussion Status
The discussion is ongoing, with participants providing insights and questioning assumptions. Some guidance has been offered regarding the need to clarify the problem statement and the components involved. There is a lack of consensus on the correctness of the equations presented, and further clarification is sought.
Contextual Notes
Participants note the absence of certain forces in the provided diagram and express confusion regarding the setup of the problem. There are indications of missing information and a need for clearer definitions of the problem context.
- #2
voko
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Forces and accelerations are vectors. Once you have Newton's second law in the vector form, you obtain equations for the components of forces and accelerations. In this case, you only care about the horizontal components. What are they?
- #3
hav0c
- 58
- 0
horizontal components
of acc. are-- a and (nsinθ)/m
and force are-- ma and nsinθ
but my main question is that is my equation 3 correct?
of acc. are-- a and (nsinθ)/m
and force are-- ma and nsinθ
but my main question is that is my equation 3 correct?
- #4
voko
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hav0c said:
It is not that "a and (nsinθ)/m" are "acceleration components". a = (nsinθ)/m is the horizontal acceleration component.
I do not see any "g" in the picture so based solely on the picture, that equation is definitely wrong.
- #5
hav0c
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I have taken g to be the acceleration downwards due to gravityvoko said:
- #6
voko
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Perhaps you need to describe your problem in full detail. What you have now makes little sense.
- #7
- #8
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As are the diagrams. Please state the question exactly as it was given to you.hav0c said:
- #9
hav0c
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I wasnt having any actual problem in solving questions it was just that i wanted to check whether or not i could derive relations by using different x-y directions, but the new general equations turned out to be useless.