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?

Question

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?

anonymous · Answer

attachment

anonymous · Answer

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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

anonymous · Answer

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!

anonymous · Answer

Thanks @AllTehMaffs

anonymous · Answer

^_^

anonymous · Answer

hmmm - there should be a negative sign there I think.  blah :P

anonymous · Answer

Aha no it's correct no worries it worked out to be the right answer

anonymous · Answer

hooray! ^_^