View Full Version : Help with a vector problem
motionman04
Sep18-04, 10:44 PM
Find the horizontal and vertical components of the d = 140 m displacement of a superhero who flies from the top of a tall building following the path shown in Fig. P3.18 where = 35.0°.
The picture is here http://www.webassign.net/pse/p3-20alt.gif
Not sure of exactly where to start.
Start by drawing a triangle and writing what you know on it.
motionman04
Sep18-04, 10:52 PM
well i was able to get the vertical by 140 cos 35, but as I tried 140 sin 35, the answer turned out to be wrong.
Remember your trigonometry: what does cos 35 equal in this triangle? What does sin 35 equal?
rocky811
Sep18-04, 11:04 PM
I have a question....do you know what the right answer is...because i think i figured it out......ALSO, I don't know if I'm looking at the problem right...but to me the VERTICAL component (Y) looks to me to be opposite of the angle, which would make the y component 140 sin 35...and the x component is adjacent to the angle, which is 140cos 35
That is what I was getting at, hoping motionman would figure it out on his own.
motionman04
Sep18-04, 11:17 PM
Well I know that sin = opposite over hypotenuse, and cos = adjacent over hypotenuse, I tried 140 sin 35 but apparently thats wrong
The vertical (y) component should be 140 sin 35, the horizontal (x) component should be 140 cos 35. Make sure your calculator is in degree mode.
motionman04
Sep18-04, 11:30 PM
yep, its in degree mode, and I get an answer of 80.3007 m, however, its wrong for some reason
CinderBlockFist
Sep19-04, 12:53 AM
140sin35 = 80.3007, so u are right, what does the answer say in the book?
motionman04
Sep19-04, 01:05 AM
Well its a problem is on webassign.com, and each time I put in that number, it says it is incorrect
CinderBlockFist
Sep19-04, 01:12 AM
maybe put negative -80.3007, since it is negative according to the x and y axis on how the diagram is labeled. Other than that, the only thing i can think of is check your significant digits, and how accurate webassign.com wants ur answeres to be.
motionman04
Sep19-04, 01:13 AM
Yep, that did the trick, thanks for that one cinderblock
CinderBlockFist
Sep19-04, 01:15 AM
No problem bro
motionman04
Sep19-04, 01:17 AM
I've got another problem thats been stumping me, it goes like this:
Instructions for finding a buried treasure include the following: Go 77.0 paces at 245°, turn to 130° and walk 110 paces, then travel 100 paces at 162°. The angles are measured counterclockwise from an axis pointing to the east, the +x direction. Determine the resultant displacement from the starting point.
Now, I've found the x, y components for all three vectors, by doing
77 cos/sin 245, 110 cos/sin 130, 100 cos/sin 162, and after finding the resultant vector for the x, it turned out to be wrong.
Pyrrhus
Sep19-04, 01:17 AM
I hate this Web applets, if you put too many numbers it will say wrong...
try putting -80.30
Pyrrhus
Sep19-04, 01:19 AM
I've got another problem thats been stumping me, it goes like this:
Instructions for finding a buried treasure include the following: Go 77.0 paces at 245°, turn to 130° and walk 110 paces, then travel 100 paces at 162°. The angles are measured counterclockwise from an axis pointing to the east, the +x direction. Determine the resultant displacement from the starting point.
Now, I've found the x, y components for all three vectors, by doing
77 cos/sin 245, 110 cos/sin 130, 100 cos/sin 162, and after finding the resultant vector for the x, it turned out to be wrong.
Use the components!!
Sum all the X components and the Y components, and you will have the displacement in X component and Y component.
Well other than that, let me try
\vec{R} = (R_{x}i + R{y}j) m
R_{x} = 77cos(245) + 110cos(130) + 100cos(162)
R_{y} = 77sin(245) + 110sin(130) + 100sin(162)
\vec{R} = (-198.35i + 45.38j) m
|\vec{R}| = \sqrt{(-198.35)^2 + (45.38)^2} = 203.47 m
\theta_{R} = 167.11^o Counter-Clockwise.
motionman04
Sep19-04, 01:25 AM
Yeah, I did, I got the resultant vector for the x components and used that as the displacement vector from the starting point
Pyrrhus
Sep19-04, 01:26 AM
Yeah, I did, I got the resultant vector for the x components and used that as the displacement vector from the starting point
i don't think that's what is asking. It's asking for the Resultant Vector.
motionman04
Sep19-04, 01:40 AM
haha no luck with the 221 or 315 degrees,
Pyrrhus
Sep19-04, 01:42 AM
haha no luck with the 221 or 315 degrees,
That's the answer... :grumpy:
It's a simple polygone shape with one side being the Resultant vector.....
Pyrrhus
Sep19-04, 01:44 AM
Copy me the exact problem with the underlines, so i can see what's expecting...
motionman04
Sep19-04, 01:52 AM
Well when I added the y resultant, it was -123.148
motionman04
Sep19-04, 01:56 AM
Wait nevermind, i got 45.38 for the y
Pyrrhus
Sep19-04, 01:57 AM
Wait nevermind, i got 45.38 for the y
you're correct, i probably put some value wrong in the calculator...
Pyrrhus
Sep19-04, 02:10 AM
so did the darn thing, finally accepted the answer?
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