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I know the stack exchange does not help on hw problems, but please understand that I am just self learning these topics, I'm not in a class, and I'm using my dads old engineering mechanics book... Now my book says I'm completely wrong on a VERY simple vector problem (Question 2.6). This bothers me and want to see if you guys come up with their answer of: 18.95°.

My Answer: 52.02°.

enter image description here

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    $\begingroup$ We have no problem helping you with homework questions. However, we ask that you edit your question showing your work. What did you do to get your answer? That way we can better focus on what you're doing wrong and therefore need to improve on. $\endgroup$
    – Wasabi
    Aug 15 '16 at 19:16
  • $\begingroup$ @Wasabi, ok thank you for letting me know that, and i will be sure to post my work on future problems! $\endgroup$
    – chris360
    Aug 15 '16 at 19:17
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    $\begingroup$ @chris360 You can also extend your this question by editing it. If you do this before the close vote is ready, it will have a significant chance to survive this vote. Another important thing: scanned homework questions look very bad, votes in border cases tend to be much more lenient if you show also you are working on it. Only to send a scanned paper and then copy-paste the answer into your homework, this is what very unpopular here and I think it is right. $\endgroup$
    – peterh
    Aug 15 '16 at 20:42
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What steps did you take to try to solve the problem?

b-b is the only "output" for vertical components in force F. I'll call the force in b-b $F_{b-b}$ and the force in F $F_F$. This means that the vertical component of force in $F_{b-b}$ must equal the vertical component of force in $F_F$. That is,

$$ F_F \sin{\alpha} = F_{b-b} \sin{60^{\circ}} \\ $$ Then, just solve for $\alpha$:

$$ \sin{\alpha} = \frac{F_{b-b}}{F_F} \sin{60^{\circ}} \\ \alpha = \mbox{asin}\left({\frac{F_{b-b}}{F_F} \sin{60^{\circ}}}\right) \\ $$

Now, plug in the numbers:

$$ \alpha = \mbox{asin}\left(\frac{150}{400}\sin{60^{\circ}}\right) \\ \alpha = \mbox{asin}\left(0.375*0.866\right) \\ \alpha = \mbox{asin}(0.3248) \\ \boxed{\alpha = 0.3308 \mbox{ rad} \\ \alpha = 18.95^{\circ}} $$

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    $\begingroup$ Thank you!! I made a stupid mistake and literally just now solved it just before you posted! Thank you for your time! $\endgroup$
    – chris360
    Aug 15 '16 at 19:15
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    $\begingroup$ @chris360 - Generally speaking, when you post a question like this, you should write down all the steps you took in trying to solve the problem, with an English description of what you're attempting to accomplish with each step. You should actually do this for all your work - future homework, tests, research projects, theses, software - English comments are a lifesaver when you go back weeks or years later and try to figure out what you were doing. $\endgroup$
    – Chuck
    Aug 15 '16 at 19:22
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    $\begingroup$ Anyways, you might find that you solve your own problem in just writing the description of the problem and what you tried. I've solved my own problems like this, posting here and subsequently deleting the question before it was asked, dozens of times, sometimes on extraordinarily difficult projects I've been working on for months. $\endgroup$
    – Chuck
    Aug 15 '16 at 19:23

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