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# mECHANICS watch

1. A basic question, but....if you resolve a force in two components at right angles, the sum of the forces adds up to more than the original. Why is that??
2. Use Pythagoras to work out the total force, forces are vectors, not just magnitude.
You're effectively asking why the length of two sides of a triangle dont add up to make the hypotenuse's length!
3. Caps lock fail...

But yeah, a^2 + b^2 = c^2
4. (Original post by rezarf)
Use Pythagoras to work out the total force, forces are vectors, not just magnitude.
You're effectively asking why the length of two sides of a triangle dont add up to make the hypotenuse's length!
I sort of get that, but what I can't get round is...

If I had a force acting of 10N acting at 45 degrees to the horizontal (no gravity), I would have a vertical and horizontal component each of about 7N. If had a power station at each of these locations i would be getting a total of 14N force from an input of 10n. Obviously wrong but why?
5. (Original post by skinny100)
I sort of get that, but what I can't get round is...

If I had a force acting of 10N acting at 45 degrees to the horizontal (no gravity), I would have a vertical and horizontal component each of about 7N. If had a power station at each of these locations i would be getting a total of 14N force from an input of 10n. Obviously wrong but why?
Im kind of confused what you're asking
But because the force is acting at an angle the cannot just add them as straight lines
Vectors add tip-to-tail, they have direction
6. (Original post by skinny100)
I sort of get that, but what I can't get round is...

If I had a force acting of 10N acting at 45 degrees to the horizontal (no gravity), I would have a vertical and horizontal component each of about 7N. If had a power station at each of these locations i would be getting a total of 14N force from an input of 10n. Obviously wrong but why?
In the vein of the question.

Work done is force times distance moved in the direction of that force.

If 10 N force moves 1 m (in its direction) work done is 10 Nm.

Horizonal component of force is 10/sqrt(2) distance moved in horizontal direction is 1/sqrt(2). Work done in horizontal direction is 10/2 = 5 Nm.

Same vertially. So total work done is 5+5 = 10.
7. (Original post by rezarf)
Im kind of confused what you're asking
But because the force is acting at an angle the cannot just add them as straight lines
Vectors add tip-to-tail, they have direction
If you resolve a force up and own a slope and at right angles to it you equate the component against other forces. So in a 45 degree slope you would say that a force of 10N is acting down the slope with 7N( maybe against something pulling it back) and it is also acting against another force at 90 degrees to the slope at 7N. It might therefore be opposing two forces each of 7N . Total 14N. How is that?

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