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Please show how you did it? And the answer should be in symbols. (no numbers)
The beam is to be hoisted using three forces as shown. If the resultant force is directed along positive y axis, determine the magnitude of force FB and angle such that the magnitude of force FB is minimum in terms of other two forces and angles. The resultant force compensates the weight of the beam hoisted.
Please show how you did it? And the answer should be in symbols. (no numbers)
The beam is to be hoisted using three forces as shown. If the resultant force is directed along positive y axis, determine the magnitude of force FB and angle such that the magnitude of force FB is minimum in terms of other two forces and angles. The resultant force compensates the weight of the beam hoisted.
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Since the resultant force is directed along positive y axis, the sum of horizontal components of the 3 forces must equal 0 N.
- Fb * sin θ + Fa * sin θa + Fc * sin 60˚ = 0
Eq. #1: Fb * sin θ = Fa * sin θa + Fc * sin 60˚
The sum of the vertical components of the 3 forces must equal the weight of the beam.
Fb * cos θ + Fa * cos θa + Fc * cos 60˚ = m * g
Eq. #2: Fb * cos θ = (m * g) – (Fa * cos θa + Fc * cos 60˚)
To determine the angle, θ, divide Eq. #1 by Eq. #2
sin θ ÷ cos θ = tan θ
tan θ = (Fa * sin θa + Fc * sin 60˚) ÷ [(m * g) – (Fa * cos θa + Fc * cos 60˚)]
θ = inverse tangent of (Fa * sin θa + Fc * sin 60˚) ÷ [(m * g) – (Fa * cos θa + Fc * cos 60˚)]
Fb * sin θ = Fa * sin θa + Fc * sin 60˚
Fb * cos θ = (m * g) – (Fa * cos θa + Fc * cos 60˚)
(Fb * sin θ)^2 + (Fb * cos θ)^2 = (Fa * sin θa + Fc * sin 60˚)^2 + [(m * g) – (Fa * cos θa + Fc * cos 60˚)]^2
To determine the equation for Fb, use the following equation, sin^2 + cos^2 = 1
(Fb * sin θ)^2 + (Fb * cos θ) = Fb^2 * (sin^2 θ + cos^2 θ) = Fb^2 * 1 = Fb^2
Fb^2 = (Fa * sin θa + Fc * sin 60˚)^2 + [(m * g) – (Fa * cos θa + Fc * cos 60˚)]^2
Fb = square root of the above
Since the resultant force is directed along positive y axis, the sum of horizontal components of the 3 forces must equal 0 N.
- Fb * sin θ + Fa * sin θa + Fc * sin 60˚ = 0
Eq. #1: Fb * sin θ = Fa * sin θa + Fc * sin 60˚
The sum of the vertical components of the 3 forces must equal the weight of the beam.
Fb * cos θ + Fa * cos θa + Fc * cos 60˚ = m * g
Eq. #2: Fb * cos θ = (m * g) – (Fa * cos θa + Fc * cos 60˚)
To determine the angle, θ, divide Eq. #1 by Eq. #2
sin θ ÷ cos θ = tan θ
tan θ = (Fa * sin θa + Fc * sin 60˚) ÷ [(m * g) – (Fa * cos θa + Fc * cos 60˚)]
θ = inverse tangent of (Fa * sin θa + Fc * sin 60˚) ÷ [(m * g) – (Fa * cos θa + Fc * cos 60˚)]
Fb * sin θ = Fa * sin θa + Fc * sin 60˚
Fb * cos θ = (m * g) – (Fa * cos θa + Fc * cos 60˚)
(Fb * sin θ)^2 + (Fb * cos θ)^2 = (Fa * sin θa + Fc * sin 60˚)^2 + [(m * g) – (Fa * cos θa + Fc * cos 60˚)]^2
To determine the equation for Fb, use the following equation, sin^2 + cos^2 = 1
(Fb * sin θ)^2 + (Fb * cos θ) = Fb^2 * (sin^2 θ + cos^2 θ) = Fb^2 * 1 = Fb^2
Fb^2 = (Fa * sin θa + Fc * sin 60˚)^2 + [(m * g) – (Fa * cos θa + Fc * cos 60˚)]^2
Fb = square root of the above