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srg

Gold Member

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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 [Broken]

A horizontal force of [itex]\vec{F}=400 lbs[/itex] is placed on the structure at point

To solve this, I drew a diagram

http://srg.sdf.org/images/PF/StaticsHW2.png [Broken]

I used the Pythagorean theorem to find

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!

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

http://srg.sdf.org/images/PF/StaticsHW.png [Broken]

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 [Broken]

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!

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