# I really nead some help in my science class

1. Apr 27, 2004

### gbdavidx

I know this is my only post but i really need help...I don't really understand any of this and i am not doing well in my class ... I would be willing to pay you (only one person if you can help me out tonight with all the questions, pacific time) or you can just help me out for free and post the problem(s) with the number on these boards which is what im looking for!!!!... or if anyone has done these problems they could scan them and email them to me heres what i'm looking at, thanks in advance!!!

http://gbdavidx.net/new1.jpg [Broken]
http://gbdavidx.net/new2.jpg [Broken]

Thank You Very Much!!!!

Last edited by a moderator: May 1, 2017
2. Apr 27, 2004

### Diane_

Some help - I hope enough

There's really too much information there for me to give you any meaningful specific help without actually doing the problems for you. Let me try giving you some general guidelines.

First off, remember Newton's First and Second Laws: basically, forces cause accelerations and accelerations are caused by forces. Any time you see an object accelerating, there are forces involved somewhere. Similarly, any time forces are applied to objects, accelerations will result unless the forces are balanced by others acting in the opposite direction. (That's an oversimplification, but it's close.)

Gravity acts as a force on any object, particularly if you're talking about objects near the surface of the Earth. If you consider an object sitting on a table, for instance, there is a force due to gravity (called the "weight") pulling it down. Since we observe that the object doesn't fall, there must be a force opposing the weight. This is the force exerted by the surface of the table on the surface of the object. This force always acts perpendicular to the surface, so it is called the "normal force" - normal being mathematician-speak for "perpendicular". Any time an object is resting on a surface, you will have two forces (at least) acting - the weight and the normal force. If the surface is parallel to the surface of the Earth, the two will cancel out exactly.

When you try to move an object along a surface, there is a force between the two surfaces that will resist that motion. This force is called friction. Friction is a complicated process, but we can model it with a reasonable degree of accuracy by saying that there is some number, called the "coefficient of friction", that will give us the frictional force if we multiply it by the normal force between the two surfaces. So, for instance, if we're dealing with a 10.0 kg object sliding along a flat table at a constant speed, with a coefficient of sliding friction of .5, we know that:

1) The weight of the object is mg = (10.0 kg)(9.80 m/$s^2$ = 9.80 N.

2) The normal force between the two objects must cancel out the weight, so it's also 9.80 N.

3) The frictional force is then (.5)(9.80N) = 4.90 N.

4) Sincer the object is not accelerating ("constant speed" and direction), the net force must be 0 - i.e. the frictional force must be cancelled out by something.

5) Therefore, someone or something is pushing the object with a force of 4.90 N in the direction of motion, to cancel the 4.90 N frictional force acting in opposition.

If the object is on an incline, then you'll have to decompose the forces acting into those normal to and parallel to the surface in order to determine the normal force and frictional forces. I didn't look through the problems enough to determine if this is something you need to worry about - if you aren't dealing with inclines, then ignore this.

I hope that helps, at least some. In summary, the equations that express the laws you want to use are:

Newton's Second Law: F = ma, F and a being vectors.

Weight: W = mg, W and g being vectors. (Note that this is the same as Newton's Second Law, specifically for gravitational forces.)

Friction (sliding): F = (mu)N, mu = coefficient of sliding friction, N = normal force. Neither F nor N are vectors here, as the directions are perpendicular.

Friction (static): F < (mu)N, where mu is again the coefficient of friction but a different one from above and N is the normal force. Here, static friction acts to keep things from starting to move. It will increase from 0 to its maximum possible value, at which point the thing starts to slide and we move over into the other regime.

Let me know if this is insufficient. Good luck.

3. Apr 27, 2004

Wow, Diane. You put in a lot of work!

gbdavidx, please don't ask for help on your entire assignment... Imagine doing your own homework. And now imagine doing it another 6 or 7 times in the same night. If everybody posted their entire assignment, that's what it'd be like for those of us who try to help out. Needless to say, it's not very pleasant.

Of course, you're welcome to ask for help on a few questions that are giving you trouble or even the concepts behind them. But many of the questions are almost exactly the same as the others, so that's writing out the same explanation twice, which wastes our time and doesn't help you any.

The moral of the message: Please ask only a few questions at one time. Learn from those and apply what you learned to your other problems. And then be happy that your homework's done.

4. Apr 28, 2004

### gbdavidx

Updated

Ok I figured out most of the stuff but theres only a couple questions where i didn't understand. On 9C when it says "Describe what must happen with the forces on the ar to get the speed back up to 60 MPH", I don't know what to do and on 9A i don't understand it why the Air Resistance 1200 Pounds, I believe that was the answer but I don't understand why that is. Thanks... This time im more specific :)

5. Apr 29, 2004

### ophecleide

The acceleration on an object is always going to be in the direction of the net force. If you take your foot off the pedal, the only force acting on the car is air resistance, so the car will slow down. In order for the car to accelerate back up to 60MPH, it must accelerate and in order for it to accelerate in a given direction (in this case, forward), it must have a net force in that direction greater than the force in the other direction. The only force in that direction is the force on the tires and the only opposing force is air resistance. Does that make sense?

Remember what Diane was saying about Newton's first and second laws. If the car is not accelerating (as the problems explains) there cannot be any net force acting on it. If the force of the road pushing on the tires (I know it sounds wierd to put it that way, but that's how it works) is 1200 lbs, how much force from the air is required to balance that out?