Does the slope effect the gravitational acceleration??

[cyeokpeng]

How is the reation force at the microscopic level related to the electromagnetic force?

[cyeokpeng]

How is the reation force at the microscopic level related to the electromagnetic force?

[swansont]

According to Newton First Law' date=' there is no net force. [b']Obviously, that force acts against the gravitational force is the force by the ground. It is a reaction force which maintain your equillibrium and it is passive, so it is called normal force.[/b]

 

The stuff in bold is absolutely wrong.

 

Action and reaction forces are always of the same type. If the action force is gravitational, so is the reaction force.

 

It is important to remember that action and reaction forces act on different objects. The action is the gravitational force the earth exerts on the object, so the reaction is the gravitational force the object exerts on the earth.

 

Since the motion of an object is dependent upon the force acting on it, when you draw a free-body diagram, thre will never be an action-reaction force pair in it.

[swansont]

According to Newton First Law' date=' there is no net force. [b']Obviously, that force acts against the gravitational force is the force by the ground. It is a reaction force which maintain your equillibrium and it is passive, so it is called normal force.[/b]

 

The stuff in bold is absolutely wrong.

 

Action and reaction forces are always of the same type. If the action force is gravitational, so is the reaction force.

 

It is important to remember that action and reaction forces act on different objects. The action is the gravitational force the earth exerts on the object, so the reaction is the gravitational force the object exerts on the earth.

 

Since the motion of an object is dependent upon the force acting on it, when you draw a free-body diagram, thre will never be an action-reaction force pair in it.

[swansont]

How is the reation force at the microscopic level related to the electromagnetic force?

 

The reaction force isn't, but the normal force is. At the microscopic level, the contact force between objects is the repulsion of electrons.

[swansont]

How is the reation force at the microscopic level related to the electromagnetic force?

 

The reaction force isn't, but the normal force is. At the microscopic level, the contact force between objects is the repulsion of electrons.

[albertlee]

I asked the another teacher in another grade group,... and he told me that a force can be divided into unlimited vector components in any direction.....

 

I understand the idea by now...., but Can any one tell me how to use trig and Pythagoras theorem in calculation of components by now???

 

Secondly, please do tell me the reason why the method works.....

 

thx

 

Albert

[albertlee]

I asked the another teacher in another grade group,... and he told me that a force can be divided into unlimited vector components in any direction.....

 

I understand the idea by now...., but Can any one tell me how to use trig and Pythagoras theorem in calculation of components by now???

 

Secondly, please do tell me the reason why the method works.....

 

thx

 

Albert

[swansont]

Go here and scroll down to "rectangular components"

[swansont]

Go here and scroll down to "rectangular components"

[Primarygun]

It is important to remember that action and reaction forces act on different objects. The action is the gravitational force the earth exerts on the object, so the reaction is the gravitational force the object exerts on the earth.

Thanks for giving advice.

So the wrong thing raised by me is that " the ground exert a force", right?

Gravity pushes us touch grounds and exert forces onto grounds, then grounds exert force against that force, isn't it ?

[Primarygun]

It is important to remember that action and reaction forces act on different objects. The action is the gravitational force the earth exerts on the object, so the reaction is the gravitational force the object exerts on the earth.

Thanks for giving advice.

So the wrong thing raised by me is that " the ground exert a force", right?

Gravity pushes us touch grounds and exert forces onto grounds, then grounds exert force against that force, isn't it ?

[albertlee]

Yes PrimaryGun, the components of Fg exerting on the ground will have the same reaction back to the object.....

[albertlee]

Yes PrimaryGun, the components of Fg exerting on the ground will have the same reaction back to the object.....

[swansont]

Thanks for giving advice.

So the wrong thing raised by me is that " the ground exert a force"' date=' right?

Gravity pushes us touch grounds and exert forces onto grounds, then grounds exert force against that force, isn't it ?[/quote']

 

The ground certainly exerts a force. It's just not the reaction force to the weight - it's not gravitational in nature. It's the reaction force to the contact force the object exerts on the earth.

 

Reaction force, by Newton's third law, has a very specific definition. And is in general being misused in this thread. If we are concerned with the motion of an object, and we draw a diagram with all the "action forces" that are exerted on the object, then there will be no reaction forces in the diagram. The reaction forces are those exerted by the object. There will be one for every action force.

[swansont]

Thanks for giving advice.

So the wrong thing raised by me is that " the ground exert a force"' date=' right?

Gravity pushes us touch grounds and exert forces onto grounds, then grounds exert force against that force, isn't it ?[/quote']

 

The ground certainly exerts a force. It's just not the reaction force to the weight - it's not gravitational in nature. It's the reaction force to the contact force the object exerts on the earth.

 

Reaction force, by Newton's third law, has a very specific definition. And is in general being misused in this thread. If we are concerned with the motion of an object, and we draw a diagram with all the "action forces" that are exerted on the object, then there will be no reaction forces in the diagram. The reaction forces are those exerted by the object. There will be one for every action force.

[Primarygun]

Ya every action has reaction. In the free body diagram, reaction force is ommited unless there is a net force at that directon with the position, right?

[Primarygun]

Ya every action has reaction. In the free body diagram, reaction force is ommited unless there is a net force at that directon with the position, right?

[violetendncy]

It helps in situations like this to make a Free-Body Diagram (drawing of all the forces coming from a point mass....the dot in your diagram representing the block). It REALLY helps and simplifies everything to then rotate your drawing and look at everything as if the surface of the ramp were perfectly horizontal. This way, it is VERY clear which force is responsible for the acceleration down the ramp--the weight--, and the normal force is pointing straight up.

 

We know the magnitude of the weight force is Mg.....the mass of the block(usu. in kilograms) times the assumed and famous "g" or about 9.80665 meters/second/second.

 

If we know the height of the ramp (at its highest point, perhaps) from the PERFECTLY level ground, and if we know the length of the ramp (that is the length of the side of the ramp that is resting on the ground...the ramp's bottom), we can easily calculate the angle of inclination as the inverse tangent of the height (Y) divided by the length (X). Call it slope....the inverse tangent of "m", then (interestingly enough, "m" is used to denote slope because the French word for to rise is "moter"....if you happen to know why it came from France, I'd love to know). Now then, we have at this point a little picture of a ramp, a square box on top of it, and two vectors....two forces, that is. One is the normal force extending from the surface of the ramp directly under the block perpendicularly outward. The other is the weight force of the block extending from the block's centre of mass directly to the centre of the earth (interestingly enough, there is more mass in the southern hemisphere of the earth, but you know of course the earth we're talking about is highly idealized, etc. etc.). This weight force is attempting to pull the block through the ramp and into China. Because this is obviously NOT happening, and because the block isn't floating or flying into space, we can say that the SUM OF ALL THE FORCES IN THE VERTICAL DIRECTION IS ZERO NEWTONS (vertical meaning the up-down axis along which the normal force is acting---remember we rotated our picture to make the normal force go straight up and down....to simplify things).

 

Now to calculations....

 

We know Newton's second law as the Impulse-momentum thingy, but we can do some hand waving to come up with the more familiar F=Ma.

 

Therefore, we'll use this equation in conjuction with equations made by observation to isolate for variables that aren't so immediately obvious and solve for them.

 

Bust it up into components....

 

Fy=Forces perpendicular to the ramp surface=Ma=N-Mgcos(angle of incl.)

 

We said this equals zero.

 

Notice also that I did NOT just call the normal force Mg....only weight can be defined that way for situations like this...but as i said the Y-component of the Normal force IS Mg, indeed!

 

ergo...

 

Fy=MAy=N-Mgcos(angle of incl.)=0

thus, N=Mgcos(angle of incl.)

 

We now have an expression for the magnitude of N...the normal force. This is because the ENTIRE normal force is directed in opposition to the Y-component of the Weight force.

 

We see that there is NO acceleration in this direction (Ay=0).

 

Now make for X stuff (parallel to the surface of the ramp)...

 

Fx=MAx=Mgsin(angle of incl.)

 

Here we can simply divide-out by the mass of the block/trolley we have the acceleration down the ramp...

 

Fx/M=Ax=gsin(angle of incl.).

 

I've found that solving stuff like this by experiment is more difficult than it sounds due to the soup of CHAOS we are swimming around in. Don't let it discourage you though, because usually you can figure out what's messing you up the most and erradicate it from everything else you do.

[coquina]

I saw an experiment in a science museum once which I think explains what you are describing.

 

A model locomotive was on a track - it had a funnel at the top of its smoke stack in which rested a red ball. The operator could push a button which launched the ball a short distance into the air. He did this several times while the train was sitting still, then asked the audience (Mostly children) what they thought would happen with the train started moving. The general concensus was that the ball would fall behind the train. However - the ball kept right on dropping into the funnel. He launched the ball just before the train went under a bridge - the train went under, the ball went over, and the ball fell right back in the funnel.

[ecoli]

That won't always happen. If shot high enough, the ball will not land in the funnel, due to the rotation of the earth. Also why a pendulum will never stay swinging exactly straight.

[YT2095]

that is kinda bringing in external forces outside the frame this example was made in though

 

in a word "irrelevant"

[hughw]

sorry for being not understandable...' date=' but this is my first time dealing with the change of acceleration due to the inclined angle....

 

This is an useful site: http://lectureonline.cl.msu.edu/~mmp/applist/si/plane.htm

 

but also where I need explanation from...

 

for the red arrow, there is no question, it is the force due to gravity....

 

but what I dont get are those two black arrows, why the force due to gravity change its direction for those?? I mean, the gravitational force is always downwards, if it can go in those directions, can the gravity force goes upwards direction????

 

such confuse.....

 

Any body can help??

 

Albert[/quote']

 

Just a Side note:

in the example "Simple Inclined Plane", it is only real-world valid for phi< 45^O, Any angle larger and the cube will tend to tip rather than slide down the plan.

[Klaynos]

Please, find an older post to bounce...