Moments Question: Bike Equilibrium Conditions

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SUMMARY

The discussion focuses on determining the equilibrium conditions for a bicycle with a mass of 15 kg, where the front and back wheels contact the ground at points P and Q, respectively, with a distance of 1 meter between them. The center of gravity is positioned above point O, located 30 cm from point P. The two key equations derived from the principle of moments are: 0.3R1 = 0.7R2 and R1 + R2 = 150, where R1 and R2 represent the normal reactions at the front and rear wheels. These equations satisfy the conditions for static equilibrium of the bicycle.

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Homework Statement


A bicycle of a mass 15kg resting in a vertical position, with the front and back wheels touching the horizontal ground at points P and Q respectively, where PQ=1m. The centre of gravity of the bike is vertically above O, a point on PQ where PO=30cm.

Given that the normal reactions of the ground on the front and rear where are R1 and R2, write 2 equations in R1 and R2 which satisfy the conditions of equilibrium.


The Attempt at a Solution


So I know you have to use the principle of moments to give the equations. I am just unsure where the pivot is. Is it at the line of center of gravity of the bicycle?
 
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The pivot can be at any point that you choose. Since it asks for the equation in R1 and R2, then yes take them about O. If you draw it out, you know the horizontal distances d1 and d2 from O to R1 and R2.
 
Ok, taking O as the pivot I get:

Eqn 1:
sum of CM=sum of ACM
0.3R1=0.7R2

Eqn 2:
sum of upward forces=sum of downward forces
R1 + R2=150

I'm assuming that's the answer?
 

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