Newtons second law homework

  • #1
438
0

Homework Statement



A cyclist and her bicycle have mass 75kg. she is riding on a horizontal road, and positions herself so that 60% of the normal contact force is on the back wheel and 40% on the front wheel. The coefficient of friction between the tires and the road is 0.8. What is the greatest acceleration she can hope to achieve?
Whilst riding at 6ms^(-1), she applies both brakes to stop the wheels rotating. In what distance will she come to a stop?

Homework Equations



Newtons second law

The Attempt at a Solution



I think the friction force of the back wheel will be the forward force of the bicycle so

0.48R=75a ,i dont consider the frictional force of the front wheel ?

I have no idea for the next.
 

Answers and Replies

  • #2
61
0


In what distance will she come to a stop?

You can calculate friction force of both wheel. Friction force have to do work on distance to stop. Conservation of energy is the easiest way to solve this problem.

regards
 
  • #3
438
0


You can calculate friction force of both wheel. Friction force have to do work on distance to stop. Conservation of energy is the easiest way to solve this problem.

regards

thanks but isn't the friction force provide the forward force for the bicycle?

and also is part 1 correct?
 
  • #4
61
0


thanks but isn't the friction force provide the forward force for the bicycle?
Yes, but only friction from back wheel (which is driven) provide forward force. During braking both wheels "produce" force.

I dont clearly understand what you mean by "0.48R=75a"?

regards
 
  • #5
438
0


Yes, but only friction from back wheel (which is driven) provide forward force. During braking both wheels "produce" force.

I dont clearly understand what you mean by "0.48R=75a"?

regards

For (1), the friction of back wheel, (0.8)(0.6R) provides the forward force, ma.

Therefore , (0.8)(0.6R)=ma ,also do i consider the friction of the front wheel here?

ie 0.48R-0.32R=ma ?
 
  • #6
61
0


Therefore , (0.8)(0.6R)=ma ,also do i consider the friction of the front wheel here?

ie 0.48R-0.32R=ma ?
No. When bicycle accelerated, front wheel did not produce friction. Only back wheel is driven by a cyclist.

So, in (1) you have 0.8*0.6*m*g=m*a where g is 9.81m/s^2.

BTW... I think, she can hope to achieve that acceleration, but she does not have enough power.

regards
Bartek
 
  • #7
438
0


No. When bicycle accelerated, front wheel did not produce friction. Only back wheel is driven by a cyclist.

So, in (1) you have 0.8*0.6*m*g=m*a where g is 9.81m/s^2.

BTW... I think, she can hope to achieve that acceleration, but she does not have enough power.

regards
Bartek

Thanks Bartek,

For (1) ,isn't that the front tyre is in contact with the ground,so when it rubs against the ground ,fricton is produced?

For(2), can i just use the motion formulas ,v=u+at

I don think so because the acceleration is not constant in this case?

(2)
 
  • #8
61
0


For (1) ,isn't that the front tyre is in contact with the ground,so when it rubs against the ground ,fricton is produced?

For(2), can i just use the motion formulas ,v=u+at

I don think so because the acceleration is not constant in this case?
When the bike start and accelerates friction force is propelled force. So, only force from back wheel works in this case - because only back wheel is driven by a biker. Front wheel theoretically resist against movement, but it is rolling friction - you can omit it (such as an air resistance etc.).

When bicycle is braking both wheels are stationary and rub on the road. Both of them produce friction which slow bicycle. You should use the formula for distance in uniformly delayed movement. Or you can use conservation energy law.

Why do you think that acceleration is not constans?
 
  • #9
4
0


Can you be of any help, I need a simple (if that is possible) equation to help me determine the weight increase of a 12 stone occupant of a car doing 30mph coming to a dead stop?
 
  • #10
61
0


Can you be of any help, I need a simple (if that is possible) equation to help me determine the weight increase of a 12 stone occupant of a car doing 30mph coming to a dead stop?
According https://www.physicsforums.com/showthread.php?t=5374" you have to start new thread, and show that you have attempted to answer your question in order to receive help...

Try this :-)

regards
 
Last edited by a moderator:
  • #11
438
0


When the bike start and accelerates friction force is propelled force. So, only force from back wheel works in this case - because only back wheel is driven by a biker. Front wheel theoretically resist against movement, but it is rolling friction - you can omit it (such as an air resistance etc.).

When bicycle is braking both wheels are stationary and rub on the road. Both of them produce friction which slow bicycle. You should use the formula for distance in uniformly delayed movement. Or you can use conservation energy law.

Why do you think that acceleration is not constans?

thanks for your help!
 

Related Threads on Newtons second law homework

  • Last Post
Replies
2
Views
661
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
597
  • Last Post
Replies
5
Views
4K
Top