mg =33.5•9.8 =328,3 N.
The beam is in equilibrium; therefore, net torque and net force are zero.
Clockwise torque - Counter clockwise torque =0
The pivot point is at the right end of the beam.
mg is applied at the center of the beam, which is 2.075 m from either end.
T is applied at a distance 4.15 – 1.20 = 2.95 m
T • 2.95 - 328.3•2.075 =0
T = (328.3 • 2.075)/ 2.95 = 230.9 N
Net force is zero
T↑ mg↓ F ↑
F = 328.3 - 230.9
F = 97.4 N
I would tell you the steps I've taken so far but I have no idea where to even start. Any help is appreciated! Thanks!
A uniform 33.5-kg beam of length = 4.15 m is supported by a vertical rope located d = 1.20 m from its left end as in the figure below. The right end of the beam is supported by a vertical column.
(a) Find the tension in the rope.
_______N upward
(b) Find the force that the column exerts on the right end of the beam.
_______N upward
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