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If I am touching a car chassis and it has 12V of force from the positive terminal of the battery, wouldn't it have 12V of force in the chassis if the cars system theoretically had no voltage drop? Wouldn't I get shocked with some of that 12V if I touched the chassis and earth ground? Given that my resistance is higher than the path back to the negative terminal it would be minimal? What would happen if you attached the positive lead to the chassis? Similarly, pretend I have a piece of metal attached to the negative terminal of a battery and have a wire coming from the positive terminal connecting to said piece of metal with no load. Why is it that the wire and metal get's very hot, yet when you measure the voltage between the metal and an earth ground source, the multimeter reads 0V? Is the current only trying to reach the negative lead of the battery and nothing else? If this is so then why do you get shocked from sticking something metal in the positive terminal of a wall socket while touching something metal. Isn't it the same thing or is it because one is DC and one is AC?

If I am touching a car chassis and it has 12V of force from the positive terminal of the battery, wouldn't it have 12V of force in the chassis if the cars system theoretically had no voltage drop? Wouldn't I get shocked with some of that 12V if I touched the chassis and earth ground? Given that my resistance is higher than the path back to the negative terminal it would be minimal?

The voltage difference that the car battery or alternator causes for the car is localized. It is not referenced to ground voltage.

Aside from that your body’s resistance is too high to cause electrocution. Some current may flow from the positive terminal through your body, through the ground towards the tire, then up the tire, through the car’s chassie, then through anything that might be connected to the negative post of the battery, and finally reaching the battery. Any such current will be so small that you can’t even measure it. For all practical purposes there is no current going through your body.

What would happen if you attached the positive lead to the chassis? Similarly, pretend I have a piece of metal attached to the negative terminal of a battery and have a wire coming from the poLsitive terminal connecting to said piece of metal with no load. Why is it that the wire and metal get's very hot, yet when you measure the voltage between the metal and an earth ground source, the multimeter reads 0V? Is the current only trying to reach the negative lead of the battery and nothing else?

The multimeter will read 0V because the voltage of the car and the voltage of ground are not related to each other. I think you are assuming that 0 V on the battery is equivalent to earth ground, and they are not. If the voltages are not related to each other through some physical contact then a current will not flow and what the voltmeter reads will be 0V or something completely random as it picks up stray voltages passing through the air from power lines or other sources.

If you want 0V on a car battery to be earth ground you can connect the negative terminal of the battery to a huge copper rod and shove that rod into the dirt. The battery is supposed to produce 12 volts higher on the possitive terminal than the negative terminal. Since we have connected the negative terminal to earth ground therefore the positive terminal will have a voltage of (EARTH GROUND + 12v). The negative terminal will be at (EARTH GROUND). If you took a multimeter and measured the voltage between the terminals of this battery it will still be 12V. When voltages are related all the multimeter does is take whatever it measured on the positive terminal and subtracts what it measures on the negative terminal.

Voltmeter Voltage = Positive Lead - Negative Lead

If we’re measuring across this battery then it would look like

Voltmeter Voltage = Positive Lead – Negative Lead = (12V + Vearth) - (Vearth) Voltmeter Voltage = 12V

If you flip the leads of the multimeter to make it measure backwards it’ll say -12v instead of +12V..

If this is so then why do you get shocked from sticking something metal in the positive terminal of a wall socket while touching something metal. Isn't it the same thing or is it because one is DC and one is AC?

The difference is that it is AC. AC has a far easier time traveling through the human body than DC. It also helps that the wall voltage is 120V AC instead of 120V DC. If you were to be grounded and touch 120V DC you’ll probably still be fine though it might burn your finger at the site where there current entered your body. I say probably. That’s still a pretty high voltage to be careless with, and it’s possible it’ll take a path through your heart and kill you. Aside from this the voltage that comes from a wall socket is already related to earth ground so if you’re standing on the ground it’s far easier for the current to take a path through your body. AC voltage oscillates between (120V + Earth GROUND) and (-120V +Earth Ground). As I’ve mentioned before AC also has a far easier time traveling through the human body than DC like in a car battery.

This is exactly why SIMULATING a circuit is better than actually touching things in the real world. Best to avoid touching anything electrical buddy. But if you must, use the ONE HAND RULE: Put one hand in your pocket and only use your other -- one hand -- to touch electricity, if you can't help yourself. That way you can avoid any current flow thru your body. This assumes your feet are not grounded. Be safe... simulate here instead.