Is a bicycle or dirtbike an example of f=ma?

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A dirt bike is a suitable example of Newton's F=ma principle, as its engine generates force that results in torque, ultimately leading to the bike's acceleration. The interaction between the wheel and the ground involves torque, which is the rotational equivalent of force. The point where the wheel contacts the ground is referred to as the "contact patch." The torque applied to the wheel, divided by the tire's radius, produces a backward force on the ground, which in turn generates an equal forward force, leading to acceleration. Understanding these concepts clarifies the mechanics behind how both bicycles and dirt bikes operate.
fourthlaw
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Hey all - this is my first post so I mght as well say hi -

I'm not very smart, but find everyday physics interesting. I tried to look up Newton's Laws, but I can't figure out:

Is a dirt bike or bicycle a good example of the f=ma principle?

All the info I found states that the force is a 'pushing' force, i.e, someone pushing the object. I can see the bicycle with the pedals fitting the explanation, but is a dirtbike an example of this law?

Also, is there a name for the point where the wheel pushes off the ground (propulsion? inertia?) Thanks.
 
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The interaction between the wheel and the ground is called torque. Torque is the rotational analogue of force. A dirt bike is a good example of F = m A, because the engine produces a force (in the pistons) which produces a torque (through the cams) which rotates the chain which produces a torque on a wheel which rotates the wheeel which produces and acceleration of the bike as a whole.
 
ok - I thought so! Thanks for your help.
 
fourthlaw said:
Also, is there a name for the point where the wheel pushes off the ground.
The name for the "point" is "contact patch", where the tire is in contact with the ground. The torque on the wheel, divided by the radius of the tire equals the backwards horizontal force applied to the ground. The ground "reacts" with an equal and opposite forwards force and this force / mass of (rider + vehicle) = acceleration.
 
The name for the "point" is "contact patch", where the tire is in contact with the ground. The torque on the wheel, divided by the radius of the tire equals the backwards horizontal force applied to the ground. The ground "reacts" with an equal and opposite forwards force and this force / mass of (rider + vehicle) = acceleration.
When I first read that I was like ":eek:!", but once I read it again it made sense. Thanks. :approve:
 
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