In the horizontal direction,
Nsinx-Tcosx=ma=mv^2/r
You might have got your angles wrong in the triangles so as to confuse you. The orange is what you need to focus on! Opposite directions.banked tracks.JPG
Hey with a question "a train of mass m is moving with speed v around a curve of radius r banked at an angle theta. THe normal reaction between the rails and the wheels of the train is N, the lateral thrust (up the slope) is T and g is the acceleration due to gravity". I was trying to resolve forces and in the answers the horizontal resolving should be "Nsin(theta) - Tcos(theta)= (mv^2)/r
my question is how do you know the "minus" is supposed to be there??
In the horizontal direction,
Nsinx-Tcosx=ma=mv^2/r
You might have got your angles wrong in the triangles so as to confuse you. The orange is what you need to focus on! Opposite directions.banked tracks.JPG
USYD: B Sc (Adv Maths) / (B A)? I
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