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GavinxFiasco
Group Title
The massless bar, hinged at A, is inclined at an angle θ = 46.0° and subjected to horizontal and vertical forces F1 and F2, as shown. If L1 = 4.00 m, L2 = 1.00 m, F1 = 10 N, and the beam is in static equilibrium, what is the magnitude of F2?
 11 months ago
 11 months ago
GavinxFiasco Group Title
The massless bar, hinged at A, is inclined at an angle θ = 46.0° and subjected to horizontal and vertical forces F1 and F2, as shown. If L1 = 4.00 m, L2 = 1.00 m, F1 = 10 N, and the beam is in static equilibrium, what is the magnitude of F2?
 11 months ago
 11 months ago

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AllTehMaffs Group TitleBest ResponseYou've already chosen the best response.1
dw:1383951629230:dw Only the components of the force perpendicular to the bar (phi=90, psi=90) produce torque, so sum of the torques \[\sum \tau = \big(L_1F_1\sin\theta \big)\sin\phi  L_2F_2 \cos \theta\sin\psi =0\] solve for F2 I think
 11 months ago

AllTehMaffs Group TitleBest ResponseYou've already chosen the best response.1
Rewrite \[ \sum \tau = \big ( L_1 F_1 \sin \theta \big) \sin \varphi\big ( L_2 F_2 \cos \theta \big) \sin \psi=0\] \[F_2=\frac{\big ( L_1 F_1 \sin \theta \big)}{L_2\cos\theta}\] \[F_2=F_1\tan\theta\frac{L_1}{L_2}\] makes sense; if theta=0, F2 = 0 because F1 will supply no torque. If theta=90, F1 has to be zero or the function is undefined. hooray!
 11 months ago

GavinxFiasco Group TitleBest ResponseYou've already chosen the best response.0
Thanks @AllTehMaffs
 11 months ago

AllTehMaffs Group TitleBest ResponseYou've already chosen the best response.1
hmmm  there should be a negative sign there I think. blah :P
 11 months ago

GavinxFiasco Group TitleBest ResponseYou've already chosen the best response.0
Aha no it's correct no worries it worked out to be the right answer
 11 months ago

AllTehMaffs Group TitleBest ResponseYou've already chosen the best response.1
hooray! ^_^
 11 months ago
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