Question

A bicycle rider has mass M. The wheels of the bicycle are at distance l apart and horizontal distance between the seat and the rear wheel is d. Using principle of virtual work find the normal reactions on the base of the two wheels when the rider is riding steadily.​

Answer 1

Answer:

not coming.cant answer

Answer 2

Answer:, $r=\frac{md}{I-m}$

Given:

A bicycle rider with a mass $m$, he sets a distance $d$ from the rear wheel and the distance between both wheels is $l$

Explanation:

Let's first start with the free body diagram of the downward force of $m$. We have the rear wheel, force up and the front wheel reaction force up. So, start with our $3$ equations.

Some of the forces in the $x \;axis=0$ and

some of the forces on the $y\;axis=0$ and

some of the moments about-

let's say the real well about $r=0$.

So for the $x \;axis$,

we can ignore this because there is no force in the $x$ direction for the $y$axis.

Let's go right here. Some of the forces $y=r+ f-m=0\rightarrow(1)$.

For equation $2$,

some of the moments around $r=d$, and this is a distance $l$.

So you take the force times the length away from the rotational point and clockwise is our positive direction.

So the weight of the mass $m$ is going counter clockwise around the point $r$.

So it's a negative m d and then the force $f$ is going counter clockwise around the point $r$.

So it's a positive plus, $fl=0$ and now using these $2$ equations. We can solve out for at the rear force and the front forces.

So, let's start off with the front force from this bottom equation. Equation: 2, say: $fl =m d$ on t.

Therefore, $f = \frac{m d}{l}$  call this the plug.

This back into equation- $\frac{dr + md}{ l - m} = 0$.

Therefore, $r =\frac{m d}{l-m}$, and these are your $2$ equations.

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