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@sweetburger Please help me!!!! :)
Can anyone please help me find the resultant angle?!?!?!?
@MAEMAEHOCKEY final got your tagname :D
give me a couple of mins
np, take your time :)
i thought it would be simple
and i thought ......nvm. i'll just beg at someone's else's door. -_-
i willl help u i am happy to
ok i think i got it
Hey don't bother, i don't wanna make you go through stress just for a question...leave it, i'll just ask someone else...Chill...:"D
what .......i am not stressed
i'm just saying if your here to be a chill bird, then be a chill bird, i ain't trapping you in a cage. :"D
i am not
Aait then.... i just wanna let you know that i've already done 90% of the question, i just need to find the resultant angle. and Thanks!!!
ok i think i got it if i am doing it right
24 right ???
can you confirm if the 24 and 32 degrees as shown in the marked up diagram? |dw:1442883620661:dw|
hold up let me attach a picture of the actual question....don't go anywhere...promise me!!! O_O
|dw:1442883835366:dw| Yes, this is a little better, with the force magnitudes. One suggestion I make for your future solutions is to draw the forces in the same orientation as the original. This way, there is a lesser chance of making mistakes.
|dw:1442883951868:dw| So R^2=15^2+12^2-2(15)(12)cos(124)=23.8812 as you had it.
|dw:1442884257124:dw| sin(\(\theta\)/12=sin(124)/R so sin(\(\theta\)=12 sin(124)/23.8812=0.416582 \(\theta\)= asin(0.416582)=24.619 degrees (measured from the 15N force, or 24.619-24=0.619 degrees from the axis of the direction of the vehicle.
correction: a little confusion for R. Should have written R^2=......., and R=sqrt(...)=23.8812 N
@YumYum247 Still there? To find the angle, I used the sine rule, but forgot to indicate it.
yes i'm still here....!!!
0.61deg.....but that's not even a one whole degree tho :O
so you used... SineA/a = SineC/c SinA/15 = Sin124/24
That's with respect to the axis of the vehicle. |dw:1442885722739:dw|
Yes, but sin(theta)/12 = sin(124)/R. Sine rule sometimes does not give the correct angles when the unknown angle is near 90 degrees (angle or supplement). But here we know the unknown angle is acute, so there's no problem.
yah sorry i realized that.... :D
Again thank you infinite :"D
You're welcome! :)