Solving Torque Problem: Find Forces FT, Horizontal & Vertical on Beam

[xregina12]

A floodlight with a mass of 40 kg is used to
illuminate the parking lot in front of a library.
The floodlight is supported at the end of a
horizontal beam that is hinged to a vertical
pole, as shown. A cable thatmakes an angle of
11◦ with the beam is attached to the pole to
help support the floodlight. Assume the mass
of the beam is negligible when compared with
the mass of the floodlight.
The acceleration of gravity is 9.81 m/s2 .

a) Find the force FT provided by the cable.
Answer in units of N.
2056 N
b) Find the horizontal force exerted on the
beam by the pole. Answer in units of N.

c) Find the vertical force exerted on the beam
by the pole. Answer in units of N.


I got a and b but for some reason I can't get c.
My work.
Fynet=0=Tsin11+Fy-mg where Fy is the vertical force exerted on the beam by the pole.
Tsin11+Fy=mg
392.4 +Fy=392.4
Fy=0
however, that doesn't work
anyone has suggestions for part c?

 

[asleight]

A floodlight with a mass of 40 kg is used to
illuminate the parking lot in front of a library.
The floodlight is supported at the end of a
horizontal beam that is hinged to a vertical
pole, as shown. A cable thatmakes an angle of
11◦ with the beam is attached to the pole to
help support the floodlight. Assume the mass
of the beam is negligible when compared with
the mass of the floodlight.
The acceleration of gravity is 9.81 m/s2 .

a) Find the force FT provided by the cable.
Answer in units of N.
2056 N
b) Find the horizontal force exerted on the
beam by the pole. Answer in units of N.

c) Find the vertical force exerted on the beam
by the pole. Answer in units of N.


I got a and b but for some reason I can't get c.
My work.
Fynet=0=Tsin11+Fy-mg where Fy is the vertical force exerted on the beam by the pole.
Tsin11+Fy=mg
392.4 +Fy=392.4
Fy=0
however, that doesn't work
anyone has suggestions for part c?

$$\sum\tau=\tau_1+\tau_2=r\vec{w}+r\vec{T}\rightarrow\vec{w}=-\vec{T}\sin(11\pi/180)\rightarrow\vec{T}=2057N$$.

Then, we notice that $$\sin(11\pi/180)\cdot2057=392.5$$ and $$\vec{w}=392.4$$. So, maybe it's just that 0.10N that they're worrying about? IDK.