Beam hinged to wall with a cable (torque, etc)

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SUMMARY

The discussion focuses on calculating the total force exerted by a pin on a uniform beam supported by a cable. The beam has a mass of 48 kg, a length of 38 m, and is positioned at an angle of 24 degrees with respect to the horizontal. The equations of equilibrium used include the sum of forces in the horizontal and vertical directions, as well as the torque equation, which was corrected to account for the beam's center of mass. The final torque equation is correctly stated as ∑τ = sin(θ) L T - cos(θ) (L/2) mg = 0.

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lizzyb
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Hi. I've tried to answer this questions four times and so far none of my answers have been correct. Any help is greatly appreciated!

A uniform 48 kg (m) beam at an angle of 24 degrees (theta) with respect to the horizontal has length of 38 m (L). It is supported by a pin and horizontal cable. The acceleration of gravity is 9.8 m/s^2 (g). What is the magnitude of the total force exerted by the pin on the beam?

So far:

\sum F_x = F_h - T = 0 horizontal component of pin - tension of cable
\sum F_y = F_v - mg = 0 vertical component of pin - mass of beam * gravity
\sum \tau = \sin \theta L T - \cos \theta L m g = 0

Does this look ok to you? For the torque, I'm trying to use the moment arm times the magnitude of the force.

Thank you.
 
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The last equation should have been:
\sum \tau = \sin \theta L T - \cos \theta \frac{L}{2} m g = 0
and that took care of it.
 

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