# Help with Trig for Resultant Forces!

Discussion in 'Science' started by electro-web, Aug 25, 2011.

1. ### electro-webNew commenter

I feel really dim asking this but here we go...

I'm an NQT teaching Physics and I need to teach some A level stuff too. It's 15 years since I did any A level Physics and the concepts are all fine but I've forgotten some of the maths.

Could someone explain this question to me step by step, please? Basically I'm struggling with trigonometry for non-right angled triangles?

Q - The resultant forces is 60N, acting at an angle of 30 degrees to one of the forces which is 45N. Find the magnitude of the second force.

The solution is here: http://imageshack.us/photo/my-images/694/screenshot20110825at131.png/

I get that we're looking at the total horizontal and vertical components, which are a combination of the 45N force and the components of the resultant.

what I can't remember, is why, with the resultant force the horizontal component is 60 cos 30 and the vertical component is 60 sin 30.

Really stupid question, and I know I'll kick myself as soon as someone reminds me....

2. ### electro-webNew commenter

I feel really dim asking this but here we go...

I'm an NQT teaching Physics and I need to teach some A level stuff too. It's 15 years since I did any A level Physics and the concepts are all fine but I've forgotten some of the maths.

Could someone explain this question to me step by step, please? Basically I'm struggling with trigonometry for non-right angled triangles?

Q - The resultant forces is 60N, acting at an angle of 30 degrees to one of the forces which is 45N. Find the magnitude of the second force.

The solution is here: http://imageshack.us/photo/my-images/694/screenshot20110825at131.png/

I get that we're looking at the total horizontal and vertical components, which are a combination of the 45N force and the components of the resultant.

what I can't remember, is why, with the resultant force the horizontal component is 60 cos 30 and the vertical component is 60 sin 30.

Really stupid question, and I know I'll kick myself as soon as someone reminds me....

3. ### dkaranaNew commenter

Because in a right triangle sin =Perpendicular/hypotenuse and cos = base /hypotenuse.Whichever way your triangle is the result will be the same as the base and perpendicular will change accordingly.
Hope it helps

4. ### robbywilliams66

Hi
The solution you have uploaded on imageshack does not look correct.
I have constructed a correct solution for you and have it as a WORD document - if you would like to give me an email address to send it to (by private message) I'd be happy to.

6. ### electro-webNew commenter

The problem is that it's not a right-angled triangle!

I'm not entirely sure it is correct, I managed to do it another way

7. ### dkaranaNew commenter

Hi electroweb,
The mistake is in your diagram or may be I did not get it.
You need to drop perpendicular from the top vertex.The base of your new triangle will be 45 +x,hypotenuse as 60N.
Therefore the new perpendicular = 60 sin 60=30N
The base= 45+x= 52 (using pythagorous)
and x= 7
The third side of your original triangle squared = (30x30) + (7x7) (Pythagorous again)
= 30.7 N

Hope it makes sense

8. ### robbywilliams66

Hi
I've sent another WORD document by email to you with a second way of solving this problem.
You will notice that both methods give the answer as 30.797N ( 31N to two significant figures).
This second method only requires you to use sine, tan and pythagoras (once) --- whereas the first method required the cosine rule.
Hope they help - enjoy
Robby

9. ### robbywilliams66

But of course the very simplist way of doing this calculation is to use the cosine rule directly on the original triangle of forces
this gives
F(sqrd) = 60(sqrd) + 45(sqrd) - [2 x 60 x 45 x cos 30]