EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 12, Problem 59P
To determine

Mass of the object suspended from left end

Expert Solution & Answer
Check Mark

Answer to Problem 59P

  1.8 kg

Explanation of Solution

Given data:

Length of the rigid beam, L=2.5m

Spring constant, k =1250 N/m

At equilibrium, beam makes an angle θ = 17.5o with the horizontal

Height of post =1.25m

Formula Used:

For translational and rotational equilibrium:

  F=0τ=0

Calculation:

  EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 12, Problem 59P , additional homework tip  1

  EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 12, Problem 59P , additional homework tip  2

  τ=0

  mg(12Lcosθ)Fspring(12Lsinφ)=0.....(1)

From Hooke’s law

  Fspring=kΔlspring

Equation (1) becomes

  mg(12Lcosθ)(kΔlspring)(12Lsinφ)=0.....(2)

From figure (b) the relation between φ and θ is:

  φ=tan1((LLcosθ)/2(L+Lsinθ)/2)=tan1(1cosθ1+sinθ)

Substitute the value of φ in (2) :

  mg(12Lcosθ)(kΔlspring)(12Lsin(tan1(1cosθ1+sinθ)))=0.....(2)

  mg(12Lcosθ)(kΔlspring)(12Lsin(tan1(1cosθ1+sinθ)))m=(kΔlspring)(sin(tan1(1cosθ1+sinθ)))gcosθ..........(3)

  Δlspring=lstretchedlunstretched=lstretched12L..........(4)

From figure:

  2α+θ=πα=πθ2

  β=α+π2=πθ2+π2=πθ2

Apply the law of cosines to the triangle

  lstreatched2=(L2)2+(Lsin(θ2)2)2(L2)(Lsinθ2)cos(πθ2)=L24+L2sin2(θ2)+L22(2sin(θ2)cos(θ2))=L24+L2sin2(θ2)+L22(sin(2θ2))=L24+L2(1cosθ2)+L22sinθlstreatched2=L24(32(cosθsinθ))lstreatched=L2(32(cosθsinθ))

From equation (4)

  Δlspring=L2(32(cosθsinθ))12L=L2((32(cosθsinθ))1)

From (3)

  m=(kL1(32(cosθsinθ)1))(sin(tan1(1cosθ1+sinθ)))gcosθ

Substitute the values and solve:

  m=(1250N/m)(2.5m)((32(cos17.5osin17.5o))1)(sin(tan1(1cos17.5o1+sin17.5o)))2(9.8m/s2)cos17.5o=(1250N/m)(2.5m)(0.301)(0.0355)2(9.8m/s2)cos17.5o=1.786kg=1.8kg

Conclusion:

Mass of the object is 1.8 kg .

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