# Calculating Horizontal Force of a Bicycle Tire

• Fusilli_Jerry89
In summary, the conversation discusses how to calculate the horizontal force that can be exerted on the ground by the rear tire of a bicycle with specific dimensions and a maximum vertical force exerted by a rider. The conversation includes steps and equations for determining the torque of the pedal and chain, as well as the resulting force from the chain on the rear sprocket. It is determined that the horizontal force applied by the ground on the perimeter of the rear tire is equal and opposite to the force applied by the tire on the ground.
Fusilli_Jerry89

## Homework Statement

A bicycle has 30cm radius tires with gear and pedal dimensions as shown. IF the maximum vertical force that a 70kg rider can exert is equal to his/her body weight, what is the horizontal force that can be exerted on the ground by the rear tire?
http://img48.imageshack.us/img48/8492/28mn1.png

## Homework Equations

same as Spool question

## The Attempt at a Solution

well first I got 686 N [70x9.8] but I have no idea what point to use as the pivot.

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Assume it's in equilibrium. Calculate the torque of the pedal at its radius to the crank axle. Then the reverse torque of the chain at the front sprocket has to be equal and opposite, right? Then calculate the resulting torque from the chain on the rear sprocket, and that has to be equal and opposite to the torque from the tire about the rear axle, right?

So first i would go:
(686)(0.18) is that just the torque of the crank axil, or is there more to do?

First you would use units in all of your equations. And yes, so far so good. Now what force from the chain balances that torque?

would that be 123.48N? but in the opposite direction

Help me out here, son. Work out the full solution with units and annotations, and I'll check it. I honestly don't have enough time to work through each of six steps in a row checking your work all the way and trying to figure out how you calculated each intermediate step.

Sorry to have to push back, but the PF is a valuable resource for us all, and you need to learn to use it efficiently. You can do it, I know you can. Work out the full problem using units, and post your full work with some description. I'll be happy to efficiently check your work. I bet you get right!

K but just one thing, would you count the chain as one part, or would u put the torque in both sides of the chain, or duz it even matter?

I think it duz matter. A chain is a linear force transfer device. It generally transfers a force to different diameter sprockets, as in this problem. The tension force out one side of it equals the tension force out the other side. That's how you get different torques -- by having different diameter sprockets connected by the same chain).

K I learn best by seeing what some one else does. SO would it be easier if you posted your work and then I could simply see how you did it?

Fusilli_Jerry89 said:
K I learn best by seeing what some one else does. SO would it be easier if you posted your work and then I could simply see how you did it?

You seem to be missing the point. This is not a lecture session. This is a place for you to bring your work and get help when we see where you are going wrong.

Yeab but, if I cannot progress further into the question, doesn't that mean you can help me out? Maybe I don'tunderstand your help very well because I cannot interprate english as well as I can interpret math and concepts. It is not a lecture because I already know what torque is and I get the basic principles of this question, I just need help putting 2 and 2 together.

Fusilli_Jerry89 said:
Yeab but, if I cannot progress further into the question, doesn't that mean you can help me out? Maybe I don'tunderstand your help very well because I cannot interprate english as well as I can interpret math and concepts. It is not a lecture because I already know what torque is and I get the basic principles of this question, I just need help putting 2 and 2 together.

Let's break it down. The pedal is not accelerating, so the chain is applying a torque opposite the torque from the vertical force from the foot. Calculate the force from the chain on the sprocket.

The force from the chain is a tension in the top part of the chain. The bottom of the chain is slack. (or the differentnce in tension between the top and bottom of the chain provides the torque. It really does not matter how you look at it. Assume the bottom is slack) The tension in the top part of the chain applies a torque to the rear sprocket. Calculate this torque. (The chain is not accelerating, so the tension is uniform throughout the top.)

The wheel is not accelerating. The horizontal force from the ground on the perimeter of the wheel is applying a torque opposite the torque from the chain. Calculate this force.

By Newton's third law, the horizontal force the ground applies to the tire is equal and opposite to the horizontal force the tire applies to the ground. The problem is solved.

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Ok I appreciate you trying to help me, I see how frustrated you probably are. It makes more sense now that I realized the bottom part of the chain is slack (duh). K what I first did with my new found knowledge was: (686N)(18) = 12348 Nm. I then divided this by 12(the radius of the small gear) to get the tension which gave me 1029 N for the tension. Now i multiplied 1029N by 6(the radius of the pin region in the middle of the tire) which gives you the torque of that are(I hope). I divided this by 30(the tires radius) which should have given me the force of the tire pushing on the ground and vice versa. I hope this is right, though I wouldn't doubt it if it's not. Sorry, It's just I am learning on my own and it's hard to get help with things without a teacher.

Fusilli_Jerry89 said:
Ok I appreciate you trying to help me, I see how frustrated you probably are. It makes more sense now that I realized the bottom part of the chain is slack (duh). K what I first did with my new found knowledge was: (686N)(18) = 12348 Nm. I then divided this by 12(the radius of the small gear) to get the tension which gave me 1029 N for the tension. Now i multiplied 1029N by 6(the radius of the pin region in the middle of the tire) which gives you the torque of that are(I hope). I divided this by 30(the tires radius) which should have given me the force of the tire pushing on the ground and vice versa. I hope this is right, though I wouldn't doubt it if it's not. Sorry, It's just I am learning on my own and it's hard to get help with things without a teacher.

I did not check all your numbers, but it sounds like you've got it. It's just a matter of equating torques as you work through the system from one end to the other.

Hi. Thank you so much for posting this thread which I am so interested in!
I this the right way to find the answer:
earth's gravity*rider's mass*18m*(1/12)m = horizontal force*30m*(1/6)m
(9.8ms2)(70kg)(18m) / 12m = horizontal force (30m) / 6m
205N = horizontal force

brianinbwangju said:
Hi. Thank you so much for posting this thread which I am so interested in!
I this the right way to find the answer:
earth's gravity*rider's mass*18m*(1/12)m = horizontal force*30m*(1/6)m
(9.8ms2)(70kg)(18m) / 12m = horizontal force (30m) / 6m
205N = horizontal force

Your last m on each side of the first equation belongs inside the parenteheses, and all of the m should be cm, but since all those m units divide out, you finished OK.

## What is the formula for calculating the horizontal force of a bicycle tire?

The formula for calculating the horizontal force of a bicycle tire is F = μmg, where F is the horizontal force, μ is the coefficient of friction, m is the mass of the rider and bicycle, and g is the acceleration due to gravity.

## How do I determine the coefficient of friction for a bicycle tire?

The coefficient of friction for a bicycle tire can be determined by conducting a controlled experiment where a force is applied to the tire and the resulting acceleration is measured. The coefficient of friction can then be calculated using the formula μ = a/g, where μ is the coefficient of friction, a is the measured acceleration, and g is the acceleration due to gravity.

## What factors can affect the horizontal force of a bicycle tire?

The horizontal force of a bicycle tire can be affected by several factors, including the surface on which the tire is riding, the type and condition of the tire, the weight and distribution of the rider and bicycle, and the speed at which the bicycle is traveling.

## How can I increase the horizontal force of my bicycle tire?

The horizontal force of a bicycle tire can be increased by choosing a tire with a higher coefficient of friction, maintaining proper tire pressure, and distributing weight evenly on the bicycle. Keeping the tires clean and using a smoother surface to ride on can also help increase the horizontal force.

## What is the significance of calculating the horizontal force of a bicycle tire?

Calculating the horizontal force of a bicycle tire is important for understanding the dynamics and performance of a bicycle. It can also help determine the amount of force needed to accelerate or maintain a certain speed while riding, and can aid in selecting the most suitable tires for different riding conditions.

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