## anonymous one year ago Two forces with magnitudes of 25 and 30 pounds act on an object at angles of 10° and 100° respectively. Find the direction and magnitude of the resultant force. Round to two decimal places in all intermediate steps and in your final answer.

1. anonymous

Resolve the force vectors into horizontal and vertical components first |dw:1439676785822:dw|

2. anonymous

$F_x=F \cos \theta$ $F_y=F \sin \theta$

3. anonymous

Then add the components in each direction to get the resultant force in that direction. The resultant force is (from the Pythagorean theorem) $|F|=\sqrt{F_{x}^2+F_{y}^2}$ and the angle is $\theta = \tan^{-1} \frac{ F_y }{ F_x }$

4. anonymous

so confused

5. anonymous

Each of the forces makes a right triangle. The given force magnitudes are the hypotenuses of the triangles. You need to use the angles (10° and 100°) and the magnitudes to find the horizontal and vertical sides. |dw:1439677744565:dw|

6. anonymous

For example the 25 lb force acts at 10°. The horizontal component is $F_x=25 \cos 10°$ and the vertical is $F_y=25 \sin 10°$ Do the same thing to find the components of the 30 lb force

7. anonymous

fx= 30cos100deg fy=30sin100deg

8. anonymous

Now add the 2 Fx's and the 2 Fy's to get the components of the resultant force.

9. anonymous

Fx = 25 cos 10° + 30 cos 100° Do Fy

10. anonymous

fy=25sin10+30sin100

11. anonymous

Use this to find the magnitude $|F|=\sqrt{F_{x}^2+F_{y}^2}$ and this to find the direction $\theta = \tan^{-1} \frac{ F_y }{ F_x }$

12. anonymous

ohh okay thank you!

13. anonymous

you're welcome

14. anonymous

Why am I not getting any of my answer choices? What am I doing wrong?

15. anonymous

@peachpi

16. anonymous

what did you get?

17. anonymous

IFI=410.66 and theta=16.47

18. anonymous

I got Fx = 19.41 N and Fy = 33.89 N $|F|=\sqrt{19.41^2+33.89^2}=39.05~N$ $\theta=\arctan \left( \frac{ 33.89 }{ 19.41 } \right)=60.2°$