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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?
 5 months ago
 5 months ago
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?
 5 months ago
 5 months ago

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AllTehMaffsBest 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
 5 months ago

AllTehMaffsBest 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!
 5 months ago

GavinxFiascoBest ResponseYou've already chosen the best response.0
Thanks @AllTehMaffs
 5 months ago

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

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