# Mechanics question

1. Feb 10, 2004

### jimmy p

Im doing Mechanics module 2 at college (the only one!) and while working through an exercise i got stumped by this question.

"A motorcycle has a mass of 250kg, and it's rider whose mass is 80kg. The centre of mass of the motorcycle lies on a vertical line midway between its wheels. When the rider is on the motorcycle, his centre of mass is 1m behind the front wheel. Find the vertical reaction forces acting through the front and rear wheels when:

(i) the rider is not on the motorcycle.
(ii) the rider is on the motorcycle."

Then it has a stupid picture of a man on a motobike with the distance 1.4m between the centres of the two wheels (so the centre of mass is as 0.7m)

Now i looked in the back of the book after many failed attempts and the answers are:

(i) 1225N for front wheel, 1225N for rear wheel
(ii) 1449N for front wheel, 1785N for real wheel.

EDIT take gravity as 9.8 instead of 10.

2. Feb 10, 2004

### Chi Meson

I am assuming you are doing rotational equilibrium, is that right?

So balance the torques: choose the rear (draw it on the left) wheel as the point of rotation. Your clockwise torques are from the bike that is acting at its c-o-m, and the guy on the bike. The only ccw torque is from the front wheel. What force at the front wheel will produce a torque equal to the sum of the other torques?

ps: I get a total cw torque of 2028.6 Nm for the second part.

Last edited: Feb 10, 2004
3. Feb 10, 2004

### HallsofIvy

Staff Emeritus
Well, you certainly ought to be able to get

"(i) 1225N for front wheel, 1225N for rear wheel"!

With the man not on the motorcycle, the weight of the motorcycle, 250kg* 9.8= 2450 Newtons is evenly divided between front and rear wheel: 2450/2= 1225 Newtons.

The man himself has weight 80kg*9.8= 784 Newtons. He is 1 meter from the front wheel so, choosing the front wheel as "center of rotation", his that would be a torque of 784 N-m counter clockwise. That must be offset by the same torque from the force on the rear wheel: its distance from the front wheel is 1.4 m so, taking F as the force on the rear wheel, 1.4F= 784 or F= 784/1.4= 560 N. Of course, the front wheel supports the rest of his weight: 784- 560= 224 Newtons.

Adding in the 1225 Newtons on each wheel due to the motorcycle itself, The front wheel is uspporting 1225+224= 1449 Newtons and the rear wheel 1225+ 560= 1785 N as advertised!

4. Feb 10, 2004

### jimmy p

*slaps forehead* Man im a dumbass, i kept converting mass into newtons so i was getting 24500!!! what a loser....

Cheers HallsOfIvy, and thanks for your answer to Chi Meson, maybe i should explain better in future!