1. Chap 11.20

2. 11. 20 A nonuniform beam 4. 50 m long and weighing 1
11.20 A nonuniform beam 4.50 m long and weighing 1.00 kN makes an angle of 25° below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 3.00 m farther down the beam and perpendicular to it. The center of gravity of the beam is 2.00 m down the beam from the pivot. Lighting equipment exerts a 5.00-kN downward force on the lower left end of the beam.
Find the tension T in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot.
Start by sketching a free-body diagram of the beam

3. Forces on the Beam Gravity (of Beam) Cable Weight of Lamp Pivot
Horizontal Vertical
Gravity (weight of) Beam −1kN
Cable (tension)  C − C cos65° C sin65°
Weight of Lamp − 5kN
Pivot Px Py
Horizontal Equation: − C cos65°+ 0 + Px = 0
Vertical Equation: − 1kN + C sin65° − 5kN + Py = 0

4. Torque Torque (fulcrum is Pivot) = F x d = F ∙ d⊥
Gravity (of Beam) kN ∙ 2cos25°
Cable − C ∙ 3
Weight of Lamp kN ∙ cos25°
Pivot distance is
Equation: kN ∙ 2cos25° − 3C + 5kN ∙ 4.5cos25° = 0

5. Solving Torque Eq: + 1kN ∙ 2cos25° − 3C + 5kN ∙ 4.5cos25° + 0 = 0
(2kN+22.5kN) ∙ cos25° = 3C
C = 24.5kN cos25° / 3 = N x .9063/3 = N
Horizontal Eq: − C cos65°+ 0 + Px = 0
Px = C cos65°
= 7401 * = N
Vertical Eq: − 1kN + C sin65° − 5kN + Py = 0
Py = 6kN − C sin65°
= 6000N – .9063*7401 N
= 6000N – N = – 708 N

6. Another Look