Flatbed truck going round a circular road on an incline

AI Thread Summary
The discussion focuses on solving a physics problem involving a flatbed truck on an inclined circular road. The user is struggling with deriving equations for normal force (N) and frictional force (Fn) from the initial equations provided. After some back-and-forth, it is clarified that the user made an error in carrying over terms correctly during the algebraic manipulation. The importance of proper notation in LaTeX for trigonometric functions is also highlighted. Ultimately, the user resolves their confusion by identifying the mistake in their calculations.
whdahl
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Homework Statement


The problem statement is in the attachments.

Homework Equations


F=ma

The Attempt at a Solution



I uploaded the solution to the problem. I have had no trouble deriving the first two equations for the summation of forces in the x, n and t directions, however I am having trouble understanding the algebra involved in the solution, specifically where it says, "Solve the first two equations for N and Fn to obtain:", then it lists the equations for N and Fn. I can't figure out how they derived those two equations for N and Fn from the first two equations.
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Solve the first two equations for N and Fn to obtain:", then it lists the equations for N and Fn. I can't figure out how they derived those two equations for N
... they did exactly what was described - they too the first two equations... these were:
$$N\cos(10^\circ)-F_{n'}\sin(10^\circ) - mg = 0\\
F_{n'}\cos(10^\circ) + N\sin(10^\circ) = m\frac{2t^2}{30}$$ ... do you not know how to solve simultaneous equations?
 
I know how to solve simultaneous equations, but I am not getting the solution they show in this case. Solving one of the equations for Fn or N and substituting that in the second equation will allow you to eliminate one of the variables. However I am not getting the solutions they provide, I am getting jumbles of sin squared's over cosines and tangents...
 
Last edited:
Please show your working.
 
I don't see where my mistake is, but I've ended up with an extra cos(10) in the final answer.
 

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Literally the moment I sent that, I found my mistake. I did not carry the \frac{mgsin10}{cos10} over correctly.
 
Well done.
Note: if you put a backslash before the name of the trig function in LaTeX, it will typeset properly.
$$\frac{mg\sin\theta}{\cos\theta}=mg\tan\theta$$
 
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