Hortizontal acceleration of a motorcycle

[BananaRed]

I have trouble starting on this problem can someone please help me?

The distance between the centers of the wheels of a motorcycle is 155cm. The center of mass of the motorcycle, including the biker, is 88.0 cm above the ground and halfway between the wheels. Assume the mass of each wheel is small compared to the body of the motorcycle. The engine drives the rear wheel only. What horizontal acceleration of the motorcycle will make the front wheel rise off the ground?

 

[cookiemonster]

Not only is the driving force of the wheel providing horizontal acceleration, it's also providing torque. Calculate the torque necessary to life the bike off the ground and go from there.

cookiemonster

 

[Soilwork]

Hey man I asked exactly the same questions a little while back.
Just go back and check the titles.
I think I called it wheelies?
something like that
But yes it has to do with the torque of horizontal acceleration and the torque of the weight of the bike.

 

[ShawnD]

What you have to do is compare the moment created from gravity with the moment created from the wheels pushing on the ground. For a moment, the force must be perpendicular to the distance. The moment created by the wheels must equal the moment created by gravity. Now write a formula for the moment at each of those (since you don't want to deal with both at the same time).

First take the moment around where the wheel touches the ground. Since gravity acts straight down, we need the horizontal distance from the centre of mass to the centre of the back wheel.
Now take the moment around the centre of mass. Since the force on the ground from the wheels is horizontal, we need the vertical distance.
Since the sum of them is 0, they are both equal.

mgd = Fh

start subbing in other variables

mgd = mah

mass is both sides so let's just forget it

gd = ah

isolate a (acceleration)

a = \frac{gd}{h}

g is gravity (9.8 m/s^2), d is distance from centre of mass to the back wheel (0.775m), and h is the height of the centre of mass (0.88m)

a = \frac{(9.8)(0.775)}{0.88}

a = 8.63 \frac{m}{s^2}