Torque and force on a tire iron

[Jason DiCaprio]

So I was curious about something. Having a flat tire today and having to change my tire made me wonder. I understand torque and force are 2 different things though related they have separate meanings.

So my question is if I have a 2 foot tire iron and I am torquing my nuts. Let's say they reached maximum torque and I can't move the tire iron anymore. Now let's say I am applying 10 pounds of force at 1 foot from the nut. My question is will I be adding 10 pounds of weight to the car/front wheel If I can't torque anymore and the nut is as tight as possible.

What if I was applying 10 pounds at 2 feet from the nut, common sense tells me I should be adding the same amount of weight to the car regardless if I apply the weight on the tire iron 1 foot away , 2 feet or any amount of feet. Weight should be weight. But I understand the torque will be increasing on the nut as I move farther and farther away from the nut.

My question is how is it possible to increase the torque on the nut but have the same amount of weight pushing down on the nut(net weight being added to the car). What if I had some ridiculously long tire iron and I was 10 feet away pushing down 10 pounds the added weight to the car should still be the car + tire iron + plus the downward force that I'm applying to the tire iron which is 10 pounds, but then I would have a ridiculous amount of torque on the bolt.

So how is this possible for the weight/force applied to the weight of the car always being car + tire iron + 10 pounds force I am applying at any location on the tire iron but the torque will change so drastically on the nut depending on where I apply the pressure. It makes no sense to me how you can have such a variation of torque , but when the bolt is as tight as possible the extra torque from the leverage all turns into the same weight/force pushing down no matter where it is on the lever?

 

[A.T.]

Didn't you ask the same question today?
https://www.physicsforums.com/threads/torque-and-force-question.928020/

So how is this possible...

Per definition.

 

[jerromyjon]

So how is this possible for the weight/force applied to the weight of the car always being car + tire iron + 10 pounds force I am applying at any location on the tire iron but the torque will change so drastically on the nut depending on where I apply the pressure.

Suppose you have a fixed nut and tire iron on a scale with a zeroed reading and you apply a torque of 10 foot pounds on the nut. Would it matter where the contact point of the scale is?

 

[jartsa]

This is a fiendishly difficult problem

Well let's see ... let's say you do not exert a torque on the wrench, but a downwards force. The bolt exerts an upwards force on the other end of the wrench. Let's call those two forces and the distance between those two forces a torque.

Let's say the wrench arm is a pipe, you stick a screwdriver into the pipe. Now when you push down the screwdriver handle, the screwdriver exerts a torque on the wrench. Let's define the torque as the upwards force that the farthest end of the screwdriver exerts on the pipe, and the equal downwards force that the part of the screwdriver nearest to the handle exerts on the pipe, and the distance of those two forces.

Now we could say those two forces are the 'extra forces' that were gained by the use of the extension arm, I mean the screwdriver.

You see, those two forces that are responsible of the torque on the wrench arm are not pushing the wrench arm down, so the force that pushes the arm down exists in addition to the torque on the arm.

 

[jerromyjon]

Now we could say those two forces are the 'extra forces' that were gained by the use of the extension arm, I mean the screwdriver.

There can be only "one" net force...

 

[jack action]

My question is will I be adding 10 pounds of weight to the car/front wheel If I can't torque anymore and the nut is as tight as possible.

Actually, you will be adding 10 pounds of weight to the entire car, not necessarily the front axle.

Say the wheelbase of your car is 100" and your tire iron is 20", pointing toward the rear axle. Applying 10 lb pushing downward, the weight on the rear axle will increase by 2 lb (= 10 * 20 / 100) and the weight on the front axle will increase by 8 lb (= 10 - 2).

But if the tire iron is pointing in front of the car, still applying 10 lb pushing downward, the weight on the rear axle will decrease by 2 lb (= - 10 * 20 / 100; negative because the torque is in the other direction) and the weight on the front axle will increase by 12 lb (= 10 - (-2)).

As you can see, if the length of the tire iron is long enough, you will be able to lift one of the axle of the ground with a small force of 10 lb. At that point, the entire weight of the car will be on one axle + 10 lb and the weight on the other axle will be, obviously, zero.

 

[CWatters]

The car can also loose 10lb if you aren't worried about your back.