Questions about Newton's third law

[Doc Al]

It's quite common to measure the "quantity" of something using either mass or weight, mixing units at will. It's OK in everyday usage, but not acceptable when doing physics.

Using standard units, a kilogram is a unit of mass. Near the Earth's surface that kilogram has a weight of about 9.8 Newtons, which is equivalent to about 2.2 pounds (force).

 

[Byrgg]

2.2 lbs, force? Huh? I thought lbs for mass, what just happened?

 

[Hootenanny]

2.2 lbs, force? Huh? I thought lbs for mass, what just happened?

Pounds(lbs) is a unit of mass. However, pounds-force(lbf) is a unit of force or weight. One pound force is approximately 4 Newton's.

 

[Doc Al]

The pound is used (confusingly, perhaps) as a measure both of force and of mass.

The standard pound (avoirdupois pounds) is a unit of mass defined in terms of the kilogram. (look it up)

But it's also commonly used as a unit of force in the imperial system of units, where mass is in slugs. The weight of one slug is (w = mg) 32 pounds (force).

Don't get hung up on this unit conversion trivia!

Personally, I stick with SI units.

 

[Byrgg]

Pound-force? News to me, I've never heard of that, I guess that's what's meant when people use pounds when talking about pressure?

Approximately 4N is 1 lbf? How is that calculated?

1N = 1kg*m/s^2 I think... what calculation is used for getting lbf?

 

[Doc Al]

Pound-force? News to me, I've never heard of that, I guess that's what's meant when people use pounds when talking about pressure?

Right. I'd say in physics or engineering it's more common to use pounds to measure force rather than mass. But you will almost always be able to tell by the context.

Approximately 4N is 1 lbf? How is that calculated?

I don't know what you mean by "calculated". This is a conversion factor between different systems of units. More accurately, 1 pound = 4.448 N.

1N = 1kg*m/s^2 I think... what calculation is used for getting lbf?

That's a definition. An analogous one for pounds-force would be:
1lbf = 1slug*ft/s^2.

 

[Byrgg]

Oh, it uses feet as well, that's probably another thing I forgot to consider... so with ft and slug, the conversions work about to be about 4.448N? I think I've got it now

 

[Hootenanny]

Oh, it uses feet as well, that's probably another thing I forgot to consider... so with ft and slug, the conversions work about to be about 4.448N? I think I've got it now

Yeah, that is correct. You have to use imperial units for acceleration.

 

[rcgldr]

To estimate the mass of air invovled, you need to know the amount of drag force and the average velocity of the air (assuming a no wind condition here) around a car, when the car is traveling at a specific velocity.

The amount of power due to drag is equal to the velocity of the car times the drag force of the air. For example, 1 horspower = speed in mph times force in lbs divided by 375 (conversion factor).

Using 1 second as a time factor, the work done by drag equals drag force times the distance the car moves in 1 second.

The amount of work done in 1 second will equal the amount of increase in energy, mostly kinetic, of the air in 1 second.

This is the tough part, you need to meausre or estimate the average velocity of the air affected by the car. Once this is estimated, then you can calculate the mass using the relationship between kinetic energy and mass e = 1/2 m v^2.

As an example, a 2006 Z06 Corvette, which has fairly low coeffiecient of drag, takes about 450 rear wheel horsepower to go 198mph. To make things easy, I'll assume that rolling resistance consumes about 20 hp, and that drag consumes about 430hp. This translates into a drag force of about 814lbs.

At 198 mph, the car is moving 290 feet / second. With a force of 814 lbs, the energy increase of the air is 236000 ft lbs. Assume that the average velocity of the diverted air is 30mph = 44 ft/sec, then the diverted mass per second is 244 lb mass per second or 118 cubic yards (2.06 lbs / cubic yard at sea level) of air per second. If the average velocity of the diverted air was 20mph = 29.3 ft / sec, then the diverted mass would be 550 lb mass per second or 267 cubic yards of air per second.

Need a second input here to confirm my math.

 

[rcgldr]

An update to my previous post. How much of the air should be considered as diverted by a car passing through it? If you include the air that increased in velocity by 1/1000th of a mph, then the amount of involved is much larger than if you only considered air that was increased in velocity by 1mph.

Somewhat related is this picture of the Thrust SST creating an enormous shock wave. Since this is super sonic (or near super sonic) speeds, the way the air is affected is different than speeds well below supersonic, but it's a good example of how much air is affected in this case:

thrust1.jpg

 

[thiotimoline]

I'm not questioning his law or anything, I just need to understand a few things about it. I understand the basic concept of the law, for every force, there is an equal force in the oppostite direction.

But what provides the reaction force in the following situations?

A car driving, has an applied force forward, but what's the reaction force backwards? Also, in the same situation, what about friction? It applies a force backward, so what's the reaction force forward provided here?

Second, a person lifting weights, they apply a force upwards to lift the weight, what is the reaction force here provided by?

A car driving, has an applied force forward, but what's the reaction force backwards? Also, in the same situation, what about friction? It applies a force backward, so what's the reaction force forward provided here?

The reaction force is just the air resistance by air acting on the surface area of the moving car in the opposite direction to that of the car's motion as well as ground friction acting on the car's wheel, also in the opposite direction to that of the car's motion.

Second, a person lifting weights, they apply a force upwards to lift the weight, what is the reaction force here provided by?

Your hand is doing work by lifting the weights upwards. Since weight of the weights is acting downwards, the reaction force comes from your hand's upward movement when lifting the weights.