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.

 

[thomate1]

For part c, you can use the equation for torque, which is given by T = F x d, where T is the torque, F is the force, and d is the distance from the pivot point. In this case, the pivot point is at the hinge of the beam and the vertical pole.

So, we can set up the equation as follows:

T = Fy x d

Since the beam is hinged to the vertical pole, the distance (d) from the pivot point to the point where the vertical force is exerted is the length of the beam (L). Therefore, we can rewrite the equation as:

T = Fy x L

Now, we can substitute the known values into the equation:

T = (40 kg x 9.81 m/s^2) x L

Since the mass of the beam is negligible, we can ignore it in the calculation. Also, we know that the torque (T) is equal to the force provided by the cable (FT) multiplied by the distance from the pivot point to the point where the cable is attached. This distance is given by L x sin11. So, we can rewrite the equation as:

FT x L x sin11 = (40 kg x 9.81 m/s^2) x L

Solving for FT, we get:

FT = (40 kg x 9.81 m/s^2) / sin11

FT = 392.4 / sin11

FT = 2056 N

Now, we can plug this value into the previous equation to solve for Fy:

2056 N x L x sin11 = (40 kg x 9.81 m/s^2) x L

Fy = (40 kg x 9.81 m/s^2) / sin11

Fy = 392.4 / sin11

Fy = 2056 N

Therefore, the vertical force exerted on the beam by the pole is also 2056 N. This makes sense because the beam is in equilibrium, so the sum of the forces in the vertical direction must be equal to zero. Therefore, the vertical force exerted by the pole must be equal to the weight of the floodlight (mg) plus the vertical component of the force provided by the cable (Tsin11).