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Physics - resolving forces

I have a problem with a question that is about resolving vectors - i keep getting the answer that is literally impossible, as i get a triangle of sides 19.3, 19.3 and 0.2 - which im sure is wrong.

The question is:
Three forces are acting on a particle, which is originally at the origin O.
A force of 10N is acting at an angle of 60 degrees to the positive x - direction
A force of 14N is acting in the y direction.
A third force of 6N is acting at -150 degrees to the positive x - direction.
Resolve the forces, find the resultant force - its magnitude and direction.


Thanks to anyone who attempts to help me!
32169537_207757300015044_4699695047988215808_n.jpg32.7KB
Resolved vertically upwards:

14 + 10sin60 - 6sin30 = 0


Horizontally to the right:

10cos60 - 6cos30 = 0

This gives a length of 19.66 vertically upwards and 0.20 to the left (since was a negative when resolved to the right)

Use this in pythagoras' theorem to get a resultant force of 19.66N

For the direction since you know the adjacent and opposite you can work out the angle using tan(x)= o/a
so (x) = tan^-1 (0.2/19.66)
(x) = 0.58 degrees
So the resultant is 19.66N acting at 0.58degrees to the negative x-direction

I'm really working out the same as you... I don't know if its right or wrong, but I got 19.66 not 19.33 (may be a mistake)
(edited 5 years ago)
Reply 2
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Treich00
OP
9
Oh yeah, thanks for spotting my mistake with the 19.33 - i need to correct that.
So even if the 0.2 to the left and 19.66 vertically was correct - would the angle equal to tan-1(19.66/0.2)?
It kind of makes sense since the horizontal force is really small compared to the vertical one, but i just want to make sure.
Original post by Treich00
Oh yeah, thanks for spotting my mistake with the 19.33 - i need to correct that.
So even if the 0.2 to the left and 19.66 vertically was correct - would the angle equal to tan-1(19.66/0.2)?
It kind of makes sense since the horizontal force is really small compared to the vertical one, but i just want to make sure.


For the direction since you know the adjacent and opposite you can work out the angle using tan(x)= o/a
so (x) = tan^-1 (0.2/19.66)
(x) = 0.58 degrees
So the resultant is 19.66N acting at 0.58degrees to the negative x-direction


i edited it in after i realised you needed an angle
its the 0.2 over 19.66 > you probably would realise this after you did 19.66/0.2 since it would give a much larger angle, considering the force horizontal is so small the angle surely would be too.
(edited 5 years ago)

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