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neonofarm [45]
2 years ago
9

A 21.0 kg uniform beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horiz

ontal.
If the angle between the beam and the cable is θ = 66.0° what is the tension in the cable?

What is the vertical component of the force exerted by the hi.nge on the beam?
Physics
1 answer:
astra-53 [7]2 years ago
4 0

The tension in the cable is 112.64 N and the vertical component of the force exerted by the hi.nge on the beam is 93.16 N.

<h3>Tension in the cable</h3>

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (21 x 9.8)/(2 x sin66)

T = 112.64 N

<h3>Vertical component of the force</h3>

T + F = W

F = W - T

F = (9.8 x 21) - 112.64

F = 93.16 N

Thus, the tension in the cable is 112.64 N and the vertical component of the force exerted by the hi.nge on the beam is 93.16 N.

Learn more about tension here: brainly.com/question/24994188

#SPJ1

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      cos (270 + 45) = F_{3x} / F₃3

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the force can be found in each axis

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         F_{x} = F_{1x} + F_{2x} + F_{3x}

         F_{x} = 80.5 +57.77 + 92.63

         F_{x} = 230.9 N

Axis y

         F_{y} = F_{1y} + F_{2y} + F_{3y}

         F_{y} = 0 + 57.77 -92.63

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we use the Pythagorean theorem

         F = √(Fₓ² + F_{y}²

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        F = 233.52 N

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        tan θ = \frac{F_y}{F_x} }

        θ = tan⁻¹ (\frac{F_y}{F_x} })

        θ = tan⁻¹ (-34.86 / 230.9)

        θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

          θ' = 360 -θ

          θ‘= 360- 8.59

          θ' = 351.41º

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