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Question 3 in section 7.6 of the AS Text Book

Calculate the tension in the crane cable and the support forces on the girder using moments and equilibrium equations.

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Question 3 in section 7.6 of the AS Text Book

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  1. Question 3 in section 7.6 of the AS Text Book 10m 6m A crane is used to raise one end of a 15kN girder of length 10.0m off the ground. When the end of the girder is at rest 6.0m off the ground, the crane cable is perpendicular to the girder as shown. a calculate the tension in the cable b Show that the support forces on the girder from the ground has a horizontal component of 3.6kN and a vertical component of 10.2kn. Hence calculate the magnitude of the support force.

  2. 10m 6m x P 15kN • a calculate the tension in the cable • This is a straightforward moments calculation. • we assume the girder is uniform and its centre of mass is therefore at the centre. • We now calculate the moment produced by the centre of mass perpendicular to the pivot P which must be equal to the moment produced by the tension in the cable. So we need to know the component of the weight of the girder perpendicular to the pivot (x):

  3. 10m 6m θ x θ P 15kN The component of the weight of the girder perpendicular to the pivot (x): The angle θ that the girder has been turned through is shown in the diagram. sin θ = 6/10 = 0.6 and angle θ = 36.90 x=15 cos θ = 12kN θ x 15kN

  4. T 10m 12kN P Now the moments around P are taken as the beam is in equilibrium 5.0x 12.0 = 10.0 x T and T = 6.0 kN

  5. Tv T= 6.0 kN 15kN P b Show that the support forces on the girder from the ground has a horizontal component of 3.6kN and a vertical component of 10.2kn. Hence calculate the magnitude of the support force. We can see that the beam is in equilibrium so the vertical component of the tension in the cable + the vertical component of the reaction at P must equal the weight of the beam, Tv + Pv = 15kN

  6. Tv T= 6.0 kN θ 15kN P Tv + Pv = 15kN Pv = 15-Tv Tv = 6.0 cosθ =4.8 Pv= 15 – 4.8 = 10.2kN 6.0 kN Tv θ=36.7o

  7. Because the girder is stationary the horizontal forces Th and Ph have to be equal. T= 6.0 kN Th Ph P Th Ph=Th = T sinθ = 6.0 sin 36.9 = 3.6kN T θ

  8. Finally we have to find the resultant force vector P by combining the vertical and horizontal components P Pv Ph Th Ph = 3.6kN Pv = 10.2N By Pythagoras P Pv Ph P =10.8kN

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