SOLUTION: Solve for a Mgsin(theta)-f=ma; f=Ia/r^2 My answer: mg(sin(theta))/((I/r^2)+m)= a Answer supposed to be (gsin(theta))/(1+I/r^2)= a

Algebra ->  Equations -> SOLUTION: Solve for a Mgsin(theta)-f=ma; f=Ia/r^2 My answer: mg(sin(theta))/((I/r^2)+m)= a Answer supposed to be (gsin(theta))/(1+I/r^2)= a       Log On


   

Question 1015488: Solve for a
Mgsin(theta)-f=ma; f=Ia/r^2
My answer: mg(sin(theta))/((I/r^2)+m)= a
Answer supposed to be (gsin(theta))/(1+I/r^2)= a

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Mgsin(theta)-f=ma; f=Ia/r^2
Is it 2 eqn or 2?
It's not clear.
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Mgsin(theta)- Ia/r^2 = ma
Mgsin(theta) = Ia/r^2 + ma
Mgsin(theta) = a*(I/r^2 + m)
a = Mgsin(theta)/(I/r^2 + m)