1. ## statics

i am stuck on question 7 a. i am trying to resolve the forces vertical and horizontal and let forces up=forces down and forces left=forces right. i can not resolve them though as i do not know what angle the 15 N force is at.

2. ## Re: statics

I get $\theta \approx 62^\circ$

3. ## Re: statics

that is the right answer. can you please explain how you got it? thanks

4. ## Re: statics

Look at the following diagram ... see if you can work it out

5. ## Re: statics

thanks. using the cosine rule on the two different triangles and i am getting the right answer.

how can you just swap the 15N in the opposite direction like that.?

6. ## Re: statics

Originally Posted by markosheehan
thanks. using the cosine rule on the two different triangles and i am getting the right answer.

how can you just swap the 15N in the opposite direction like that.?
He did not swap it. If it is balanced, then the sum of the 5N force and 12N force must sum to the opposite of the 15N force. He just set a placeholder for what the sum of those two forces needed to be.

7. ## Re: statics

The blue vector is the sum of the 5 and 12 N magnitude vectors ... the red vector is the opposite of that sum.

Let the origin be at the initial point of both the 5 and 12 N vectors and let the 12 N vector lie on the positive x-axis

$(5\cos{\theta} + 12)^2 + (5\sin{\theta})^2 = 15^2$

$25\cos^2{\theta} + 120 \cos{\theta} + 144 + 25\sin^2{\theta} = 225$

$120\cos{\theta} = 225-144-25$

$\theta = \arccos\left(\dfrac{225-144-25}{120}\right) \approx 62^\circ$