Determining Angle with Vector Components

[srg]

Greetings women and men,

I have a problem in which I have to find an angle [itex]\phi[/itex].

http://srg.sdf.org/images/PF/StaticsHW.png

A horizontal force of [itex]\vec{F}=400 lbs[/itex] is placed on the structure at point A. Find angle [itex]\phi[/itex] to give the AB component of [itex]\vec{F}[/itex] a magnitude of 600 lbs.

To solve this, I drew a diagram
http://srg.sdf.org/images/PF/StaticsHW2.png

I used the Pythagorean theorem to find x: [tex]600^2=400^2+x^2 \rightarrow x=\sqrt{600^2-400^2} \therefore x=447.2[/tex]

Then I used the law of sines to find [itex]\phi[/itex]: [tex]\frac{447.2}{\sin{30}}=\frac{400}{\sin\phi} \rightarrow \frac{400\sin{30}}{447.2}=\sin\phi \rightarrow \arcsin{\frac{400\sin{30}}{447.2}}=\phi=26.6\deg[/tex]

However, [itex]\phi=26.6\deg[/itex] is not the same as the answer in the back of my textbook. I'm not sure where I went wrong. The correct answer in the textbook is [itex]\phi=38.3\deg[/itex].

Thanks!

 

[jedishrfu]

Where's the right angle? The pythagorean theorem is predicated on using a right triangle.

 

[srg]

Where's the right angle? The pythagorean theorem is predicated on using a right triangle.

Ah! Duh! That's why. Law of cosines it is. Can't believe I overlooked that.

Thanks!

 

[phinds]

Uh ... do you understand the Pythagorean Theorem? What kind of triangles does it apply to?

EDIT: I see that jedishrfu beat me to it.

 

[HalfLostAndSearching]

Hello,

It seems like you have made a small error in your calculations. When using the law of sines, you need to use the opposite side of the angle you are trying to find, which in this case is 600 lbs. Your equation should be:

\frac{447.2}{\sin{30}}=\frac{600}{\sin\phi} \rightarrow \frac{600\sin{30}}{447.2}=\sin\phi \rightarrow \arcsin{\frac{600\sin{30}}{447.2}}=\phi=38.3\deg

I hope this helps! Remember to always double check your calculations to ensure accuracy.