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Physics Questions About Vector |
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| Oct3-04, 03:50 PM | #52 |
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Physics Questions About Vector
(2.6,1.5) looks right;
what does your calculator say when you type in cos(30)? Multiply that number with 3. |
| Oct3-04, 04:00 PM | #53 |
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i get .8660254038
*3 = 2.598076211 How would you get the answer after that? Rx = Ax + Bx = 2.6 + 0 Ry = Ay + By = 1.5 + 3 Now I add Rx + Ry = 7.1 Answer in book is 5.2m at 60 above x-axis |
| Oct3-04, 04:07 PM | #54 |
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HAVE YOU FORGOTTEN PYTHAGORAS??????
The resultant vector is: [tex](\frac{3\sqrt{3}}{2},\frac{9}{2})[/tex] The length is therefore: [tex]\sqrt{\frac{27}{4}+\frac{81}{4}}=\frac{\sqrt{108}}{2}[/tex] |
| Oct3-04, 04:12 PM | #55 |
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how did u get the resultant vector?
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| Oct3-04, 04:25 PM | #56 |
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[tex]\cos(30)=\frac{\sqrt{3}}{2},\sin(30)=\frac{1}{2}[/tex]
(This is a rather well-known relation; you'll it later on) Hence A has coordinates [tex](\frac{3\sqrt{3}}{2},\frac{3}{2})[/tex] Summing A with B (coord. (0,3)) yields the resultant vector). |
| Oct3-04, 04:39 PM | #57 |
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ahh indeed it does.. this makes sense now :D
yes I learned cos(30) = 3/2 in precalculus the 30-60-90 triangle |
| Oct3-04, 04:42 PM | #58 |
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wohoo got B) correct how would you get the degree though?
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| Oct3-04, 04:48 PM | #59 |
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How do you think?
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| Oct3-04, 04:54 PM | #60 |
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ummm I know all the measurements of the sides yet no angles except that 30 degrees, but its on the outside
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| Oct3-04, 04:55 PM | #61 |
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But you know its horizontal and vertical coordinates, right?
How can you calculate the tangent of the angle using that info? |
| Oct3-04, 05:21 PM | #62 |
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the figure isnt a right triangle though
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| Oct3-04, 05:26 PM | #63 |
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You are to find the resultant vectur's angle to the x-axis; how can you construct a triangle in such a manner that the coordinates you've been given will help you find that angle?
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| Oct3-04, 05:31 PM | #64 |
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OOO I got it :D
extend the Line downward so the verticles of the triangle are [tex] tan^-^1 = 4.5/2.6[/tex] (0,0) --> (2.6,0) --> (2.6,4.5) |
| Oct3-04, 05:33 PM | #65 |
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You should get 60 degrees..
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| Oct3-04, 05:33 PM | #66 |
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Yep I got 60 degrees 59.9 i think. Thanks for your help :D
nice i think i got section 3.2 and 3.3 of my text book down ahah... tommo I will probably ask about projectile motion, hopefully you'll help me again. The textbook leaves out alot of information. Like it leaves out little steps. |
| Oct3-04, 05:37 PM | #67 |
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No problem..
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| Mar9-11, 11:25 PM | #68 |
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its an emergency...i really need to know this.....
can two vectors representing two different physical quantities be equal???? can force and displacement be equal vectors if they are in same direction and have same magnitude?? does'nt representing different phy. quantities makes them different vectors??? |
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