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# Forces as vectors help

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1. Hi, would anyone be able to help me out with these sort of questions please; it'd mean a lot thanks
2. The resultant force is the overall force: say you had 10N acting left and 7N acting right on an object, its resultant force would be 3N left as they cancel out. The same goes when these forces are at different angles.

Think about if you had 3N pushing up and 4N pushing left on a particle - these two lines drawn out would be perpendicular. You'd notice the particle moves towards the top left, as this is the "net force", i.e. the resultant force.

Like vectors in maths, a translation of (3,4) could be thought of as moving 3 right and 4 up, making an L shape, or it could be thought of as a direct diagonal line. This diagonal line is the resultant - hence we can make a right angled triangle and use trigonometry, by treating the forces as lengths. We end up with a magnitude of 5N (see a and b)

Once you have the magnitude, you need to find an angle. Here we can use Cos = Adjacent/Hypotenuse to get 36.9 degrees (see c)

For your question, you have 4 different forces. To deal with these, simply find the resultant of one pair, then the resultant of the other pair, then finally the resultant of the two new forces. If you struggle with it, let me know!
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3. (Original post by Ed5)
The resultant force is the overall force: say you had 10N acting left and 7N acting right on an object, its resultant force would be 3N left as they cancel out. The same goes when these forces are at different angles.

Think about if you had 3N pushing up and 4N pushing left on a particle - these two lines drawn out would be perpendicular. You'd notice the particle moves towards the top left, as this is the "net force", i.e. the resultant force.

Like vectors in maths, a translation of (3,4) could be thought of as moving 3 right and 4 up, making an L shape, or it could be thought of as a direct diagonal line. This diagonal line is the resultant - hence we can make a right angled triangle and use trigonometry, by treating the forces as lengths. We end up with a magnitude of 5N (see a and b)

Once you have the magnitude, you need to find an angle. Here we can use Cos = Adjacent/Hypotenuse to get 36.9 degrees (see c)

For your question, you have 4 different forces. To deal with these, simply find the resultant of one pair, then the resultant of the other pair, then finally the resultant of the two new forces. If you struggle with it, let me know!
Hi, sorry for bothering you again (and thank you very much for your response) but I've completed the force of one pair, however, for the second pair i have no idea how i'd do it. Sorry if i'm acting a bit stupid on this topic but I've only just started mechanics this week :P

4. No worries, I've mostly forgotten it myself!
You may want to try find the resultant of the 8/6 and the 9/2 pairs as they look easier to work with.

The problem is that these aren't right angled triangles: to fix this, break them down into right angled triangles and go from there. If all else fails, you could use other trigonometry such as the sine rule. If you're still stuck tomorrow I'll give it a go then - too tired now

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