Mechanical Physics Question.

[Markymark]

Would it be a plane if there wasn't?

Depends whether you think the velocity of an object changes what it is known as.

 

[Piemaster]

[QUOTE 4150474, member: 9609"]And if the above two examples were the same how would you express it in torque? if for example the wheel radius was 1.5m then the torque need would be 75kg metres[/QUOTE]
You can't express both examples as a torque. Torque is rotational, a direct push is linear
direct push is Newtons (N)
torque = force x distance from centre of rotation (Nm or ft.lbs)

A 50N push on the back of the car will effectively work on the centre of the wheel, the lever to the ground is still the radius of the wheel. You would still be applying the same torque as putting a socket/lever to the centre of the wheel.

 

[Tim Hall]

You're comparing apples with oranges.

Force applied is measured in Newtons (or Kg x g if you prefer)

Torque applied is measured in Newton metres (or Kg x g x m if you prefer)

Torque is not the same as Force

Pleasingly, if you take a run of the mill eating apple, say 100g (or 4 ounces in old money), and an acceleration due to gravity of 9.8ms^-2 has a weight (force) of roughly 1 Newton.

 

[classic33]

Depends whether you think the velocity of an object changes what it is known as.

It'd not take off.

 

[Markymark]

It'd not take off.

Maybe I mis-understood your post, sorry. I am sure it didn't take off but it is still a plane.

 

[pplpilot]

If you did this in a forest and no one was around to witness it, does the car still move?

 

[PhilDawson8270]

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

Not really if you just push the vehicle, the lever is the radius of the wheel (contact with road to the axis of rotation). If the lever is longer, the force required to obtain the same torque is reduced.

Edit:
We need clarification, on whether the force required at the wheel, or the force that is imparted on the ratchet vs pushing directly?

 

[Oldfentiger]

[QUOTE 4150681, member: 9609"]so what i really need an answer to is how much torque you would need to apply to the centre of the wheel to replicate a 5 newton linear push from the back.

Im sort of thinking moving a vehicle by push at the top of the tyres may in some cases be easier or harder than pushing from the rear - you can move a tractor in a workshop by pushing at the tyre but wouldn't be able to shift it by push the rear of the tractor - and I guess the oppisite would be true of a trolley with 4" wheels[/QUOTE]
Answer: 5Nm

Edit - that's wrong.
It's 5Nm linear push at the wheel centre to replicate a 5Nm push to the rear of the car.
Torque required depends on the length of the lever

 

[Chromatic]

I reckon everyone is over-thinking this.

If you push the car at a point vertically above the axle, at the same radius as the periphery of the tyre, then the vehicle will move half the distance compared to the effort at the periphery.
Pu another way, there will be a mechanical advantage of 2:1, which is the same as a gearing reduction of 1:2.
Therefore the force required will be 25Kg.

I'm not.

 

[Fnaar]

[QUOTE 4150474, member: 9609"]On a flat level surface - if it took a force of 50kg to push a vehicle from the rear in a direct push. etc etc [/QUOTE]
It's not quite as catchy as "How much wood could a woodchuck chuck, if a woodchuck could chuck wood", but it has potential!

 

[Piemaster]

[QUOTE 4150681, member: 9609"]you can move a tractor in a workshop by pushing at the tyre but wouldn't be able to shift it by push the rear of the tractor[/QUOTE]
Thats slightly different as pushing the back of the tractor the lever will be centre of wheel to the ground - the radius. Using the top of the tyre the lever is top of wheel to ground - the diameter - so you are doubling the length of the lever.
For an idea, 5Nm is only isn't much more than finger tight on a 10mm bolt, not much at all. Think of a 1m lever with a 1/2Kg weight on the end of it - a mug of coffee should be about right.
Think the clamping bolts for the handlebars on the bikes are only around 5Nm, there won't be much choice in torque wrenches that go that low.

 

[Yellow Saddle]

[QUOTE 4150681, member: 9609"]so what i really need an answer to is how much torque you would need to apply to the centre of the wheel to replicate a 5 newton linear push from the back.[/QUOTE]


No matter how you look at the problem - from a torque to the hub idea or a linear push from the rear idea, the amount of force required to get the tractor to start moving is 500N (rounded up for sanity). Now, see the tractor as positioned about 1m from a wall and you are between the tractor and the wall. Your feet are up on the wall and you are pushing the tractor back at 500N and it starts moving. Now, put a lever of unknown length on the hub and again position yourself against the wall. How hard do you think you have to push on the lever to get the tractor moving? I'll give you a hint. The opposing force is 500N.

Edit: Tractor, not Tracktor

 

[Piemaster]

500Nm. Thats more like it. A proper torque .
That needs the ships BIG torque wrench, 3/4" drive, and it's ungainly enough to need 2 people.
Actually anything much more gets tightened hydraulically.

 

[mustang1]

If you get 6Nm of force and apply the same amount on a lever in a 37 degree direction opposing the force, the mass transit of the truck will be the net effect of the force multiplied by the length of the forward movement between the axle btwthisisbullshit so the inference you have at the beginning is all wrong.

Some people really don't know simple physics. Puh.

 

[nickyboy]

The most difficult thing about this thread is to read the nonsense being posted from a physics and mathematics perspective and not replying