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The towline exerts a force of P= 6 kN at the end of the 8-m-long crane boom. If e = 30°, determine the placement x of the hook at 8 so that this force creates a maximum moment about point 0. What is this moment?

The towline exerts a force of P= 6 kN at the end of the 8-m-long crane boom. If e = 30°, determine the placement x of the hook at 8 so that this force creates a maximum moment about point 0. What is this moment?

Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter12: Oscillatory Motion
Section: Chapter Questions
Problem 34P
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In step 4, I think we should write [ M=6x8=48kN.m ] because x=9.815m is not perpendicular to the force P.

Please correct me if I am wrong.

Thank you in advance,

+ Q&A Library P= 6 kN 8 ml'm X- Probs. 4-11/12 Question The towline exerts a force of P= 6 kN at the end of the 8-m-long crane boom. If e = 30°, determine the placementx of the hook at B so that this force creates a maximum moment about point 0. What is this moment? Ta P= 6kN Probs. 4-11/12 Expert Answer Step 1 The free body diagram of the crane is given below. Psina Step 2 Consider the moment about point O. M = Pcosa x0+Psina M = Psin a For maximumM, sina=1 a = 90° Step 3 From above figure distance x is calculated as, x = OB, + B,B = OA cos 0 + AB, tan 6 = OA cos 0+ (OA sin 0+1) tan 0 = 8cos 30° + (8sin 30° +1) tan 30° = 9.815 m Step 4 The maximum moment is calculated as, M = Px = 6 kN x9.815 m = 58.99 kN-m
Transcribed Image Text:+ Q&A Library P= 6 kN 8 ml'm X- Probs. 4-11/12 Question The towline exerts a force of P= 6 kN at the end of the 8-m-long crane boom. If e = 30°, determine the placementx of the hook at B so that this force creates a maximum moment about point 0. What is this moment? Ta P= 6kN Probs. 4-11/12 Expert Answer Step 1 The free body diagram of the crane is given below. Psina Step 2 Consider the moment about point O. M = Pcosa x0+Psina M = Psin a For maximumM, sina=1 a = 90° Step 3 From above figure distance x is calculated as, x = OB, + B,B = OA cos 0 + AB, tan 6 = OA cos 0+ (OA sin 0+1) tan 0 = 8cos 30° + (8sin 30° +1) tan 30° = 9.815 m Step 4 The maximum moment is calculated as, M = Px = 6 kN x9.815 m = 58.99 kN-m
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