Problem 7.08. A tower crane prepares to lift a 5 × 10³ kg load with its arm extended 70.0 m from the tower pivot point, a) how far from the tower should the x 104 counter weight extend in the opposite direction to ensure the tower does not topple? Assume the mass of the arms is supported by the structure and not relevant to the placement of the counterweight. b) If the counter weight is fixed at a distance of 15 m from the tower, what is the tension in a supporting cable attached from the tower to the load arm from above, at a distance of 40.0 m and an angle of 25.0°.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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**Problem 7.08.** A tower crane prepares to lift a \(5 \times 10^3 \, \text{kg}\) load with its arm extended \(70.0 \, \text{m}\) from the tower pivot point.

a) How far from the tower should the \(2 \times 10^4 \, \text{kg}\) counterweight extend in the opposite direction to ensure the tower does not topple? Assume the mass of the arms is supported by the structure and not relevant to the placement of the counterweight.

b) If the counterweight is fixed at a distance of \(15 \, \text{m}\) from the tower, what is the tension in a supporting cable attached from the tower to the load arm from above, at a distance of \(40.0 \, \text{m}\) and an angle of \(25.0^\circ\).
Transcribed Image Text:**Problem 7.08.** A tower crane prepares to lift a \(5 \times 10^3 \, \text{kg}\) load with its arm extended \(70.0 \, \text{m}\) from the tower pivot point. a) How far from the tower should the \(2 \times 10^4 \, \text{kg}\) counterweight extend in the opposite direction to ensure the tower does not topple? Assume the mass of the arms is supported by the structure and not relevant to the placement of the counterweight. b) If the counterweight is fixed at a distance of \(15 \, \text{m}\) from the tower, what is the tension in a supporting cable attached from the tower to the load arm from above, at a distance of \(40.0 \, \text{m}\) and an angle of \(25.0^\circ\).
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Given Data:

m=5000 kga=70 mM=20000 kgl=15 mθ=25o

 

 

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