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# Mechanics - Statics of a particle question watch

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1. Hi,

Ive been unable to answer this question -

Forces P, Q and R act on a particle at O in the plane of the co-ordinate axis Ox, Oy. Force P acts along Ox, Q acts along Oy, and R acts at angle Ѳ with Ox, in the first quadrant. Caculate the magnitude of the resultant force and the angle it makes with Ox when:

P=3N, Q=4N, R=5N, Ѳ=60 degrees

I know this is a long question and is one which is quite difficult to answer on a computer, but any time and help is greatly appreciated!

Thanks.
2. (Original post by malteser12345)
Hi,

Ive been unable to answer this question -

Forces P, Q and R act on a particle at O in the plane of the co-ordinate axis Ox, Oy. Force P acts along Ox, Q acts along Oy, and R acts at angle Ѳ with Ox, in the first quadrant. Caculate the magnitude of the resultant force and the angle it makes with Ox when:

P=3N, Q=4N, R=5N, Ѳ=60 degrees

I know this is a long question and is one which is quite difficult to answer on a computer, but any time and help is greatly appreciated!

Thanks.
Draw a diagram, if you haven't already; almost always a good idea with mechanics.

You need to work out the component of each force in the Ox direction, and add them all together.

Do the same for the Oy direction.

Then use pythagoras and simple trig to get the magnitude and direction.

Can you take it from there? If not, please be specific about where you are having difficulty.
3. (Original post by ghostwalker)
Draw a diagram, if you haven't already; almost always a good idea with mechanics.

You need to work out the component of each force in the Ox direction, and add them all together.

Do the same for the Oy direction.

Then use pythagoras and simple trig to get the magnitude and direction.

Can you take it from there? If not, please be specific about where you are having difficulty.
Thanks for the reply. Im just having difficulty on figuring out the components of each force, if that makes any sense?
4. (Original post by malteser12345)
Thanks for the reply. Im just having difficulty on figuring out the components of each force, if that makes any sense?
OK.

To resolve a force in a given direction.

Lets call our force F (original, I know), and say it is at 60 degrees above the horizontal.

Whats the component of the force in the horizontal direction?

Just F Cos 60

It's just the force multiplied by the cosine of the angle between the force and the direction you are resolving it in.

So if we wanted to resolve it in the vertical direction. The angle between the force and the vertical is 90-60=30.

And the vertical component would be F cos 30.

As you can see working with the cosine, if the direction is perpendicular to the force, then we'd use cos 90, which equals 0, so there is no component perpendicular to the direction of the force.

Hope that helps.

PS: There are other ways to view this using a right angle triangle, which give the same results, but in a different form.
5. (Original post by ghostwalker)
OK.

To resolve a force in a given direction.

Lets call our force F (original, I know), and say it is at 60 degrees above the horizontal.

Whats the component of the force in the horizontal direction?

Just F Cos 60

It's just the force multiplied by the cosine of the angle between the force and the direction you are resolving it in.

So if we wanted to resolve it in the vertical direction. The angle between the force and the vertical is 90-60=30.

And the vertical component would be F cos 30.

As you can see working with the cosine, if the direction is perpendicular to the force, then we'd use cos 90, which equals 0, so there is no component perpendicular to the direction of the force.

Hope that helps.

PS: There are other ways to view this using a right angle triangle, which give the same results, but in a different form.
Thanks for the help. It is really appreciated! Btw, i repped you. its the first time ive done that so i hope ive done it right!
6. (Original post by malteser12345)
Thanks for the help. It is really appreciated! Btw, i repped you. its the first time ive done that so i hope ive done it right!
Cheers, yes it came through fine.

You will also encounter the other method, which is worth mentioning. It is slightly quicker, but some people seem to get confused with it, judging by the posts on this forum.

And that is, as before the component in the direction you are resolving is F cos whatever, BUT the component perpendicular to that is F Sin (same whatever). This is just the same as if you'd taken the cos of (90-whatever), which is what I did in the previous posting.

If that made sense, fine, if not, ignore it until you encounter it.

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Updated: October 13, 2009
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