A floodlight with a mass of 50 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 that makes an angle
of 14◦ 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.
1. Find the force FT provided by the cable.
The acceleration of gravity is 9.81 m/s
2. Find the horizontal force exerted on the beam
by the pole
3.Find the vertical force exerted on the beam
by the pole. Answer in units of N.
weight = 50 g = 50 * 9.81 = 491 Newtons
force up = FT sin 14 = 491
so FT = 491/sin 14 = 2030 Newtons tension
horizontal component = 2030 cos 14 = 1970 Newtons
force up by pole = weight of 50 kg mass = 491 N if nothing accelerates
what is vertical?
1. To find the force FT provided by the cable, we can break down the forces acting on the floodlight. There are two forces acting vertically: the force of gravity and the vertical component of FT. These two forces must balance each other for the floodlight to be in equilibrium.
The weight of the floodlight is given by W = m * g, where m is the mass and g is the acceleration due to gravity. Since the floodlight has a mass of 50 kg and g = 9.81 m/s^2, the weight is W = 50 kg * 9.81 m/s^2 = 490.5 N.
The vertical component of FT is given by FTv = FT * sin(14°). So, FTv = FT * sin(14°).
Since the floodlight is in equilibrium, W is equal to FTv. Therefore, we have FT * sin(14°) = W.
Plugging in the known values, we get FT * sin(14°) = 490.5 N.
Now, we can solve for FT:
FT = 490.5 N / sin(14°).
Using a calculator, FT ≈ 2024.33 N (rounded to two decimal places).
Therefore, the force FT provided by the cable is approximately 2024.33 N.
2. To find the horizontal force exerted on the beam by the pole, we can consider the horizontal equilibrium of the floodlight. The only horizontal force acting on the beam is exerted by the pole.
Since the floodlight is in equilibrium, the horizontal component of FT must balance the horizontal force exerted by the pole on the beam.
Therefore, the horizontal force exerted on the beam by the pole is equal to FT * cos(14°). Plugging in the known value of FT, we have:
Horizontal force = FT * cos(14°) ≈ 2024.33 N * cos(14°).
Using a calculator, the horizontal force exerted on the beam by the pole is approximately 1953.63 N (rounded to two decimal places).
Therefore, the horizontal force exerted on the beam by the pole is approximately 1953.63 N.
3. Since the floodlight is in equilibrium, the vertical forces acting on the beam must also balance each other. The vertical force exerted on the beam by the pole can be calculated by summing the vertical forces acting on the beam.
The vertical force exerted on the beam by the pole consists of the vertical component of FT, which is FTv = FT * sin(14°), and the weight of the floodlight, which is W = 490.5 N.
Therefore, the vertical force exerted on the beam by the pole is given by:
Vertical force = FTv + W = (FT * sin(14°)) + 490.5 N.
Plugging in the known values, we have:
Vertical force = (2024.33 N * sin(14°)) + 490.5 N.
Using a calculator, the vertical force exerted on the beam by the pole is approximately 1063.86 N (rounded to two decimal places).
Therefore, the vertical force exerted on the beam by the pole is approximately 1063.86 N.
To find the force provided by the cable (FT), the horizontal force exerted on the beam by the pole, and the vertical force exerted on the beam by the pole, we need to break down the forces acting on the floodlight and the beam.
1. Finding the force FT provided by the cable:
First, let's assume the floodlight is in equilibrium (not rotating or moving). In this case, the sum of the forces in the horizontal direction is zero. We can use trigonometry to find the horizontal component of the force provided by the cable. Since the cable makes an angle of 14 degrees with the beam, the horizontal component (FTx) can be found using the equation:
FTx = FT * cos(14)
2. Finding the horizontal force exerted on the beam by the pole:
The horizontal force exerted by the pole (FH) is equal in magnitude but opposite in direction to the horizontal component of the force provided by the cable (FTx). This can be written as:
FH = -FTx
3. Finding the vertical force exerted on the beam by the pole:
To find the vertical force exerted on the beam by the pole (FV), we need to consider the forces acting in the vertical direction. Since the floodlight is in equilibrium, the sum of the vertical forces is zero. In this case, the vertical component of the force provided by the cable (FTy) must balance the downward force due to gravity acting on the floodlight. The vertical component can be found using the equation:
FTy = FT * sin(14)
However, since the mass of the beam is negligible compared to the mass of the floodlight, the vertical force exerted by the pole can be considered as equal in magnitude but opposite in direction to the vertical component of the force provided by the cable. So we can write:
FV = -FTy
To complete the calculations, we need to know the value of FT.