Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
11th Edition
ISBN: 9780077687342
Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 16.1, Problem 16.61P
To determine

Tension in each cable.

Expert Solution & Answer
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Answer to Problem 16.61P

The tension in cable A is 202.056lb.

The tension in cable B is 431.01lb.

Explanation of Solution

Given information:

The length of the beam is 15ft, the weight of the beam is 500lb, the deceleration of cable A is 20ft/s2, the deceleration of cable B is 2ft/s2 and distance between cable is 12ft.

The below figure represents the kinematics of the beam.

Vector Mechanics for Engineers: Dynamics, Chapter 16.1, Problem 16.61P

Figure-(1)

Write the expression of total force applied on the beam.

Ftotal=ma   ............... (I)

Here, the mass of the beam is m and the acceleration is a.

Consider the tension in the cable A is TA and the tension in the cable B is TB.

Write the expression of deceleration of cable B by using kinematics of the beam.

aB=aA+rα   ............... (II)

Here, the acceleration of cable A is aA, the distance between cable is r and the angular acceleration of the beam is α.

Write the expression of total acceleration on the beam.

a=aA+L2α   ............... (III)

Here, the acceleration of the cable A is aA and the length of the beam is L.

Write the expression of moment of inertia of beam.

I=mL212   ............... (IV)

Here, the mass of the slender rod is m and the length of the rod is L.

Write the expression of mass of the rod.

m=Wg   ............... (V)

Here, the weight of the beam is W and the acceleration due to gravity is g.

Write the expression of distance between mid of beam and the cable B as shown in Figure.

h=L2(Lr)   ............... (VI)

Write the expression of total moment about point B by applying equation of motion as shown in Figure-(1)

MB=(MB)effective   ............... (VII)

Here, the effective moment is (MG)effective.

Write the expression of moment about point B as shown in Figure-(1).

MB=TArWh   ............... (VIII)

Here, the tension in the cable A is TA and distance between mid of beam and the cable B is h.

Write the expression of effective moment about point B as shown in Figure-(1).

(MB)effective=T+Ftotalh   ............... (IX)

Here, the angular acceleration of the rod is α, the total acceleration on the beam is a and the total force is Ftotal.

Write the expression of generated torque in the beam.

T=Iα   ............... (X)

Here, the of inertia of beam is I.

Substitute Iα for T in Equation (IX).

(MB)effective=Iα+Ftotalh   ............... (XI)

Substitute TArWh for MB and Iα+Ftotalh for (MB)effective in Equation (VII).

TArWh=Iα+FtotalhTAr=Iα+Ftotalh+WhTA=1r(Iα+Ftotalh+Wh)   ............... (XII)

Substitute ma for Ftotal in Equation (XI).

TA=1r(Iα+mah+Wh)   ............... (XIII)

Write the expression of total force applied on the beam by equilibrium of the beam as shown in Figure-(1).

Write the expression of total force applied on the beam by equilibrium of the beam as shown in Figure-(1).

TA+TBW=FtotalTB=Ftotal+WTA   ............... (XIV)

Substitute ma for Ftotal in Equation (XI)

TB=ma+WTA   ............... (XV)

Calculation:

Substitute 2ft/s2 for aB, 20ft/s2 for aA and 12ft for r in Equation (II).

2ft/s2=20ft/s2+(12ft)α(2ft/s2)(20ft/s2)=(12ft)αα=18ft/s212ftα=1.5rad/s2

Substitute 20ft/s2 for aA, 15ft for L and 1.5rad/s2 for α in Equation (III).

a=(20ft/s2)+(15ft)2(1.5rad/s2)=(20ft/s2)(7.5ft)(1.5rad/s2)=(20ft/s2)(11.25ft/s2)=8.75ft/s2

Substitute 500lb for W and 32.2ft/s2 for g in Equation (V).

m=(500lb)(32.2ft/s2)=15.527lbs2/ft

Substitute 15.527lbs2/ft for m and 15ft for L in Equation (IV).

I=(15.527lbs2/ft)×(15ft)212=3493.575lbfts212=291.13lbfts2

Substitute 15ft for L and 12ft for r in Equation (VI).

h=(15ft)2((15ft)(12ft))=(7.5ft)(3ft)=4.5ft

Substitute 12ft for r, 291.13lbfts2 for I, 1.5rad/s2 for α, 15.527lbs2/ft for m, 4.5ft for h, 500lb for W and 8.75ft/s2 for a in Equation (XIII).

TA=1(12ft){(291.13lbfts2)×(1.5rad/s2)+(15.527lbs2/ft)×(8.75ft/s2)×(4.5ft)+(500lb)×(4.5ft)}=1(12ft){(436.695lbft)+(611.376lbft)+(2250lbft)}=2424.68lbft(12ft)=202.056lb

Substitute 15.527lbs2/ft for m, 500lb for W, 8.75ft/s2 for a and 202.056lb for TA in Equation (XV).

TB=(15.527lbs2/ft)×(8.75ft/s2)+(500lb)202.056lb=(133.066lb)+(500lb)(202.056lb)=431.01lb

Conclusion:

The tension in cable A is 202.056lb.

The tension in cable B is 431.01lb.

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Chapter 16 Solutions

Vector Mechanics for Engineers: Dynamics

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