Why do we feel lighter when accelerating on a bike downhill?

  • Thread starter Thread starter freswood
  • Start date Start date
AI Thread Summary
When riding a bike downhill, a cyclist feels "lighter" during acceleration due to the difference between apparent weight and actual weight. Apparent weight is influenced by the forces acting on the cyclist, such as gravity and normal force. When accelerating downhill, the net force reduces the normal force, making the cyclist feel lighter, similar to free-fall. Conversely, braking increases the normal force due to friction, causing the cyclist to feel heavier. Understanding these forces clarifies the sensations experienced during different biking scenarios.
freswood
Messages
11
Reaction score
0

Homework Statement


A guy is riding a bike down a hill. He feels "lighter" when he accelerates than when he's braking. Why?

The Attempt at a Solution


I thought that a higher normal reaction force than weight was responsible for feeling "light". But this can't be right. Is it something to do with the fact that less friction is acting on him when he's not braking?

Thanks :smile: I just need a really *basic* answer because the physics we're doing at the moment is just very basic.
 
Last edited:
Physics news on Phys.org
You are correct in thinking about reaction force. There are two things you need to consider here - apparent weight (which we feel) and actual weight. The actual weight is the downward force of gravity. Usually apparent weight and actual weight are the same, but when we are accelerating or some other force other than gravity and normal force is acting on us, the two are different. Also since this is crudely similar to free-fall (for which we feel no apparent weight). Can you now tell why he feels lighter?
 
Last edited:
freswood said:

Homework Statement


A guy is riding a bike down a hill. He feels "lighter" when he accelerates than when he's braking. Why?

The Attempt at a Solution


I thought that a higher normal reaction force than weight was responsible for feeling "light".

You are in the right direction although you have it backward. A normal force higher than the weight means that the person is feeling heavier than normal (and vice versa). Now, with a simple free body diagram it should be clear why the person feels heavier when braking (in what direction is the acceleration? That will be the direction of the *net* force, which will tell you something about the normal force exterted by the seat on the person).
 
Ah so is it because when he's braking, there is friction from the bike acting up the slope, which has a vertical component that adds to normal reaction force to give a higher upward force? Sorry, I know that's very wordy.

I hope I have this right :biggrin:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top