These results were completely unacceptable. I have dealt with this situation before and there are very systematic ways of dealing with it.

- First of all, I will share what tools are needed to remove the round nut.
- Second, I’ll review the mechanics of torque and force, so that we can better understand the issue at hand.
- Third, I’ll review a step-by-step guide that describes the twist-fit method.
- Lastly, I’ll go over some preventative maintenance guidelines and make sure you understand what the proper torque is for your car so that you can avoid the problem of overtightening.

## Tools needed

3 lb hammer, 1/2 “drive breaker bar, 1/2” drive with hex head nut extractor sockets, 1 “x36” iron tubing and new nut.

- ½ ”breaker bar $ 15-20
- 1 “diameter, 36” long iron pipe $ 10-15
- Nut / Bolt Extractor Rotary Socket Set $ 20-100
- 3 lb hammer $ 5-10
- WD-40 or alternative penetrating oil $ 5
- Replacement nut $ 3
- Total cost: $ 58-153 (if you have to buy everything)

Along with your other emergency supplies that are stored in your car, I recommend that you keep these items in the car as well, in case you need to repair your own tires while you are away from home.

## The mechanical advantage of leverage

The physics of leverage can be summed up as torque. In the context of removing a nut from a wheel, we can think of it as a simple static problem.

- Torque = rx F
- Torque = rotational force on the nut
- X = cross product
- r = length of breaker bar / lever tube
- F = applied force

I’d like to show how much torque you can generate with your body weight and compare it to the torque produced by a pneumatic impact wrench.

Impact wrenches used in auto shops range from 0-1000 + (ft-lbs). Generally, between 0 and 400 ft-lbs is most common.

Using a 24 “breaker bar and a 36” iron tube, I’ll show you how much force your body weight can produce on its own. Assuming it weighs 180 pounds, here is the torque calculation:

- Par = r X ΣF [EQ 1]
- Torque = [(24in*(1ft/12in))+(36in*(1ft/12in))] X (180 pounds at 90 ° vertical)
- Torque = 900 ft-lbs with 36 “iron pipe and 24” breaker bar;
- Torque = 360 ft-lb with 24 ”breaker bar only

Of course, additional force can be generated by jumping on the end of the pipe attached to the breaker bar. I’ll show the calculation I used to find how much torque is created from a 6 “vertical jump at the end of the pipe. To keep this calculation as simple as possible, I will consider a stopping distance equal to the thickness of the sole of a shoe combined with the estimated deflection at the lever (3 ”), and without energy losses to the environment.

During this jumping action, we can say that at the maximum height of your jump, the initial potential energy combined with the initial kinetic energy is equal to the sum of the kinetic energy and the potential energy when you are about to land at the end of The pipe.

- PHYSICAL EDUCATION
_{1} + KE_{1} = PE_{two} + KE_{two }[EQ 2]
- PE = m * g * h [EQ 3]
- KE = 1/2 * m * v
^{two} [EQ 4]
- PHYSICAL EDUCATION
_{1} = Potential energy at the peak of the jump
- KE
_{1} = Kinetic energy at the peak of the jump = 0; motionless at the top of the jump
- PHYSICAL EDUCATION
_{two} = Potential energy just before landing = 0; the height of the pipe = 0, where you are about to land
- KE
_{two} = Kinetic energy at the end of the jump.

We can say…