Well at least i think it is, it's one of those questions that you know you should be able to do, solve a few parts of it then get the wrong answer. I did actually just end up with the right answer but i'm still not sure i know why:
"A cyclist and her bicicyle have mass of 75kg. She is riding on a horizontal road and positions herself on the bike so that 60% of the normal contact force acts on the back wheel, and 40% acts on the front wheel. The coefficient of friction between the tyres and the road is 0.8
(a) What is the greatest acceleration she can hope to achieve?
(b) whilst riding at 6 metres per second, she rapidly brakes and the tyres instantly stop moving, in what distance will she come to a stop?"
For part (a) i worked out the limiting friction for the back tyre as 352.8 (0.8 x 441) then plugged that into f=ma to get an acceleration of 4.7, what i don't understand is why the force can't be greater than the limiting friction (which is what the question and answer suggest)
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Tough m1 question watch
- Thread Starter
- 11-06-2005 19:43
- Thread Starter
- 11-06-2005 20:54
(Original post by Chris.)
- 11-06-2005 21:03
i hate mechanics!!
- 11-06-2005 21:08
The friction is what moves the bike- it acts forwards. If the force pushing the back tyre around was greater than the limiting friction, the tyre would slip. It's the same with walking!
- 11-06-2005 21:12
a) well if the F= µR but also F=m1 a and R = m2 g
µm2 g = m1 a => a = (0.8 x 45 x 9.8)/75=4.704=4.7ms-2
b) since this is the maximum acceleration, it can also be the maximum deceleration right? so putting that value for a into the equation for motion
v2 = u2 +2as
s= (v2 -u2 )/2a
s= (02 -62 )/(2x-4.704)
s= 3.83m (to 3 sf)
Please correct me if im wrong, im alittle rusty with mechanics
- 11-06-2005 21:49
Is that right or am i making up my own method?
- Thread Starter
- 11-06-2005 23:19
Part (a) is correct but (b) is wrong.
I got the correct answer for (b) earlier but not 100% sure what's going on:
driving force - friction = 75a
0 - 588 = 75a
a = -7.84
then using constant acceleration formulae to get s = 2.30 metres.
I still don't really understand whats actually going on with the forces, if anybody can shed any light it would be cool, cheers.
- 11-06-2005 23:38
When she brakes, the friction slows her down (i.e. her acceleration becomes negative, a.k.a. deceleration) until she comes to a complete stop. You need to find what this acceleration is.
Applying Newton's second law (F=ma) as you did:
driving force - friction = ma --- (1)
Since there is no driving force on her now:
driving force = 0
Now, friction = uR. You're given that u=0.8, and R is the normal contact. Since there is no vertical motion going on, you have forces up = forces down from Newton's second law, i.e. R=mg. Hence:
friction = 0.8 * 75 * 9.8 = 588
Substituting this in (1) would give you the acceleration you're looking for.
Basically, you try to find F logically by considering vertical and horizontal motion seperately and applying Newton's second law, e.g.
forces to the left - forces to the right = ma (if you take left to be positive)
forces up - force down = ma (if you take take up to be positive)
If there's no motion in one of the 'planes', then the acceleration there is zero, so you can simply equate forces. e.g. if there is no vertical motion, then forces up = force down.