EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
16th Edition
ISBN: 8220100546716
Author: Katz
Publisher: CENGAGE L
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Chapter 14, Problem 27PQ

Children playing pirates have suspended a uniform wooden plank with mass 15.0 kg and length 2.50 m as shown in Figure P14.27. What is the tension in each of the three ropes when Sophia, with a mass of 23.0 kg, is made to “walk the plank” and is 1.50 m from reaching the end of the plank?

Chapter 14, Problem 27PQ, Children playing pirates have suspended a uniform wooden plank with mass 15.0 kg and length 2.50 m

FIGURE P14.27

Expert Solution & Answer
Check Mark
To determine

The tension in each of the three ropes when S with mass of 23.0kg, is made to “walk the plank” and is 1.50m from reaching the end of the plank.

Answer to Problem 27PQ

The tension in rope1 is F1=234N, tension in rope2 is F2=209N and tension in rope3 is F3=286N .

Explanation of Solution

The free body diagram is given below.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 27PQ

Here, F1 is the tension in rope1, F2 is the tension in rope2, F3 is the tension in rope 3, Fg,Child is the weight of S and Fg.Plank is the weight of Plank.

At equilibrium, the sum of forces and sum of torques must be zero.

Consider torque about an axis perpendicular to the page and through the left end of the plank.

Write the equilibrium condition for the torque about an axis perpendicular to the page and through the left end of the plank.

  Στ=0                                                                                                                      (I)

Here, Στ is the torque about an axis perpendicular to the page and through the left end of the plank.

Since forces F1 and F2 are acting at axis about which torque is to be calculated, they does not produce torque on the plank.

Write the expression for the torque due to a force.

  τ=Frsinϕ

Here, τ is the torque, F is the magnitude of force, r is the perpendicular distance between force and point of rotation and ϕ is the angle between force and position vector of point where force is acting.

The forces Fg,Child and Fg.Plank are making 90° with position vector whereas F3 makes an angle as shown in figure1.

Write expression for the net torque about an axis perpendicular to the page and through the left end of the plank.

  Στ=Fg,Childrchild+Fg.Plankrplank+(F3sinθ)r3                                                              (II)

Here, Fg,Child is the weight of S, rchild is the perpendicular distance between Fg,Child and axis , Fg.Plank is the weight of plank , rplank is the perpendicular distance between Fg.Plank and axis , F3 is the magnitude of force on rope as shown in figure, θ is the angle that F3 makes with horizontal and r3 is the distance between point where F3 acts and axis.

Write the expression for Fg,Child.

  Fg,Child=mchildg                                                                                                       (III)

Here, mchild is the mass of S and g is the acceleration due to gravity.

Write the expression for Fg,Plank.

  Fg,Plank=mPlankg                                                                                                      (IV)

Here, mPlank is the mass of Plank.

Substitute equation (III) and (IV) in (II) to get Στ.

  Στ=(mchildg)rchild+(mPlankg)rplank+(F3sinθ)r3

Substitute above equation in equation (I) to get expression for F3.

  (mchildg)rchild+(mPlankg)rplank+(F3sinθ)r3=0F3=(mchildg)rchild+(mPlankg)rplankF3sinθ            (V)

Write the equilibrium condition of the force along x direction.

  ΣFx=0                                                                                                                   (VI)

Here, ΣFx is the sum of all forces along x direction.

Write expression for net force along x direction.

  ΣFx=F1+F3cosθ                                                                                              (VII)

Here, F1 is the magnitude of tension in rope1.

Substitute (VII) in (VI) to get F1 .

  F1=F3cosθ                                                                                                         (VIII)

Write the equilibrium condition of the forces along y direction.

  ΣFy=0                                                                                                                  (IX)

Here, ΣFy is the sum of all forces along y direction.

Write net forces along y direction.

  ΣFy=F2Fg,Child(Fg,Plank)+F3sinθ                                                                    (X)

Here, F2 is the magnitude of tension in rope2.

Substitute (III) and (IV)in (X) to get ΣFy .

  ΣFy=F2mchildgmPlankg+F3sinθ                                                                      (XI)

Substitute (XI) in (IX) to get F2 .

  F2=mchildg+mPlankgF3sinθ                                                                              (XII)

Conclusion:

Substitute 23.0kg for mChild, 15.0kg for mplank , 1.00m for rchild, 1.25m for rplank, 2.50m for r3, 9.81m/s2 for g and 35.0° for θ in equation (V) to get F3 .

  F3=(23.0kg×9.81m/s2)1.00m+(15.0kg×9.81m/s2)1.25m(2.50)sin35.0°=286N

Substitute 286N for F3 and 35.0° for θ in equation (VIII) to get F1 .

  F1=(286N)cos35.0°=234N

Substitute 23.0kg for mChild, 15.0kg for mplank, 286N for F3 , 9.81m/s2 for g and 35.0° for θ in equation (XII) to get F2 .

  F2=(23.0kg×9.81m/s2)+(15.0kg×9.81m/s2)286Nsin35.0°=209N

Therefore, the tension in rope1 is F1=234N, tension in rope2 is F2=209N and tension in rope3 is F3=286N .

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Chapter 14 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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