Finding out the power by using average velocity....

[albertlee]

Again, What i really confused is still about how does it (cosine) do to the direction of the forces and distance.........in the solution of work?

 

 

Any help?

[swansont]

Again' date=' [b']What i really confused is still about how does it (cosine) do to the direction of the forces and distance.........in the solution of work?[/b]

 

 

Any help?

 

The dot product, that gives you the cosine, tells you what the projection of the vectors are onto each other. i.e. how much of the force is in the same direction as the displacement. A force that is perpendicular to the displacement can't add or remove energy, thus it does no work. A force that is in the exact same direction as the displacement does the maximum work possible. The cosine, in essence, tells you how efficiently you are doing the work with that force.

[albertlee]

Sorry Swansont, I still dont get it, you say that a force that is perpendicular does no work, but why you cannot do cos 90 then?

On the other hand, I know this is impossible because it is not in the direction of the force, but by that, if in my case(see in the picture of the 1st page in the thread), why we should do "cos"? we can just force times distance, because the distance is in the direction of the force................

 

 

Any more helps?

 

 

Really apreciate for the responds

[swansont]

Sorry Swansont' date=' I still dont get it, you say that a force that is perpendicular does no work, but [b']why you cannot do cos 90 then?[/b]

On the other hand, I know this is impossible because it is not in the direction of the force, but by that, if in my case(see in the picture of the 1st page in the thread), why we should do "cos"? we can just force times distance, because the distance is in the direction of the force................

 

The cosine tells you the component of the force doing the work. The perpendicular component doesn't contribute.

[albertlee]

Ok, Swansont, in ur previous message, the most confusing word is "component", I dont get what u r talking about component in this case.......Can u explain it?

 

How can something doing not in the direction of force still can move a distance that is not in the direction?

in my case, the trolly is moving horizontally, but the force's direction is diagonal.....

[swansont]

Ok' date=' Swansont, in ur previous message, the most confusing word is [b']"component", [/b] I dont get what u r talking about component in this case.......Can u explain it?

 

How can something doing not in the direction of force still can move a distance that is not in the direction?

in my case, the trolly is moving horizontally, but the force's direction is diagonal.....

 

A force that is at some angle is the same as the sum of two forces, one horizontal and the other vertical, and whose magnitude is given by the Pythagorean theorem if you were to draw the right triangle with the three vectors. The horizontal and vertical parts are the components. How big any given component is depends on the angle - you can use trig to figure that out.

 

In your example there are other forces present that are not named. Gravity, for instance, exerts a force downward. The upward component of the force the rope exerts acts against gravity, and makes no contribution to the forward motion. That component would be given by the sine of the angle (400N * sin30). The component of the forward force is given by the cosine (400N * cos30) (using the angle in your drawing)

 

(400*sin30)²+(400*cos30)²=400²

so you can see the component forces add (as vectors) to be the total force of 400N

[albertlee]

So..............Why the trolley does not move upwards but only forwords as the force is exerting diagonally?

[albertlee]

Any body?

[albertlee]

any body?

[swansont]

any body?

 

Prompting people when they're offline doesn't do any good.

 

The object would accelerate vertically if the vertical component of the force is greater than the weight. But that information isn't in the diagram.