Mechanical Physics Question.

[User]

 

[TVC]

I think 50kgf, the wheelnut is the fulcrum, so if you are applying the force at the radius of the wheel then the ratio is 1:1 to the point if contact with the road.

This kind of stuff makes my brain hurt.

 

[PhilDawson8270]

50kgf I reckon will be the same as the radius of the wheel. So if your ratchet was 1.5m long, and the wheel was 3m diameter. I'm going to say the torque will be the same.

 

[TVC]

Yep, when you push with 50kgf it causes an acceleration, the point of reaction against the road is the sole of your shoe. If you apply the load through the wheel then the reaction occurs where the tyre is in contact with the tarmac.

 

[nickyboy]

Force = Mass x Acceleration so, rearranging this, Acceleration = Force/Mass

Mass is constant so assuming you want to achieve the same rate of Acceleration by both methods of imparting the Force, the Force must be the same.

By the way, Force is expressed in Newtons, not Kilograms. And Torque is NewtonMetres (Nm)

 

[Tim Hall]

Force = Mass x Acceleration so, rearranging this, Acceleration = Force/Mass

Mass is constant so assuming you want to achieve the same rate of Acceleration by both methods of imparting the Force, the Force must be the same.

By the way, Force is expressed in Newtons, not Kilograms. And Torque is NewtonMetres (Nm)

Force can be expressed in kilograms. Kilograms force, kgf, as noted up thread. It's a bit of a cludge though, and means your answer won't come out in SI units.

 

[PhilDawson8270]

Force = Mass x Acceleration so, rearranging this, Acceleration = Force/Mass

Mass is constant so assuming you want to achieve the same rate of Acceleration by both methods of imparting the Force, the Force must be the same.

By the way, Force is expressed in Newtons, not Kilograms. And Torque is NewtonMetres (Nm)

I think you missed the question. It was comparing pushing a vehicle to turning it's wheel with a ratchet. If the ratchet is longer than the wheel radius then the force required to turn it will be reduced.

I still deal with some stuff also that spec kgf, I even have a torque wrench old enough that the metric scale is in kgf too

1kgf is roughly 10 Newtons

 

[Markymark]

Force can be expressed in kilograms. Kilograms force, kgf, as noted up thread. It's a bit of a cludge though, and means your answer won't come out in SI units.

Mass is not force. Mass can give force proportionate to the acceleration of the object.

F=ma

 

[Chromatic]

I'd just start the vehicle and drive it.

 

[nickyboy]

Force can be expressed in kilograms. Kilograms force, kgf, as noted up thread. It's a bit of a cludge though, and means your answer won't come out in SI units.

I'm afraid it can't.

Kg is a measure of Mass. So if you're on Earth, Jupiter, wherever...a Kg is a Kg

But force varies dependent on where you are. The Force required to lift a Kg on Jupiter is not the same as on Earth. As it does to accelerate a body

 

[Markymark]

[QUOTE 4150576, member: 9609"]just to avoid confusion the vehicle will be on planet earth. But I do take your point newtons would be is a more universal measurement.[/QUOTE]
FTFY

 

[nickyboy]

I think you missed the question. It was comparing pushing a vehicle to turning it's wheel with a ratchet. If the ratchet is longer than the wheel radius then the force required to turn it will be reduced.

I still deal with some stuff also that spec kgf, I even have a torque wrench old enough that the metric scale is in kgf too

1kgf is roughly 10 Newtons

Not quite. You can use a ratchet longer than the wheel radius of course. But if you want to accelerate the vehicle using the ratchet at the same rate as just pushing the vehicle, the forces required to be applied will be the same

 

[ColinJ]

You are basically using a lever. You can reduce the force required, but it would have to act over a bigger distance.

 

[pplpilot]

https://play.google.com/store/apps/details?id=com.autodesk.forceeffectmotion

 

[Seevio]

The force at the top of the wheel is going to be 50% of the force needed to push the car from the rear, regardless of the radius of the wheel.