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 5, Problem 47PQ

A block with mass m1 hangs from a rope that is extended over an ideal pulley and attached to a second block with mass m2 that sits on a ledge. The second block is also connected to a third block with mass m3 by a second rope that hangs over a second ideal pulley as shown in Figure P5.47. If the friction between the ledge and the second block is negligible, m1 = 3.00 kg, m2 = 5.00 kg, and m3 = 8.00 kg, find the magnitude of the tension in each rope and the acceleration of each block.

Chapter 5, Problem 47PQ, A block with mass m1 hangs from a rope that is extended over an ideal pulley and attached to a

FIGURE P5.47

Expert Solution & Answer
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To determine

What is the magnitude of tension on each rope and the acceleration on each block?

Answer to Problem 47PQ

The magnitude of tension on each rope is 38.6 N & 53.9 N and the acceleration on each block is 3.07 m/s2.

Explanation of Solution

From the given condition the 8.00 kg object will accelerate downwards, 3.00 kg object will accelerate upwards and 5.00 kg object will move to the right. As the three object connected together they all have same acceleration but the tension in each rope will vary. The free body diagram is given below.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 47PQ

Applying Newton’s laws.

For m1:

    ΣFy=FT1Fg1=m1a                                                                                (I)

Here, Fy is the force, FT1 is the tension force due to m1, m1 is the mass, Fg1 is the gravitational force on m1 and a is the acceleration.

For m2:

    ΣFx=FT2FT1=m2a                                                                       (II)

  ΣFy=FNFg2=0                                                                                (III)

Here, Fg2 is the gravitational force on m2 and FN is the normal force.

For m3:

      ΣFy=Fg3FT2=m3a                                                                             (IV)

Here, Fg3 is the gravitational force on m3 and FT2 is the tension force due to m2.

Write the equation for gravitational force.

    Fg=mg                                                                                              (V)

Here, g is the acceleration due to gravity.

Conclusion:

Using equation V find the gravitational force on each object.

    Fg1=m1g=(3.00 kg)(9.81 m/s2)=29.4 N

    Fg2=m1g=(5.00 kg)(9.81 m/s2)=49.1 N

    Fg3=m3g=(8.00 kg)(9.81 m/s2)=78.5 N

Substitute the above values in equation I, II and IV to generate another three equation with unknown factors.

    FT129.4 N=(3.00 kg)a                                                                                (VI)

    FT2FT1=(5.00 kg)a                                                                                (VII)

  78.5 NFT2=(8.00 kg)a                                                                             (VIII)

Solve the equation VI and VIII to get tension and substitute in equation VII.

    FT1=(3.00 kg)a+29.4 N                                                                            (IX)

    FT2=78.5 N(8.00 kg)a                                                                            (X)

Then equation VII becomes,

    FT2FT1=(5.00 kg)a[78.5 N(8.00 kg)a][(3.00 kg)a+29.4 N]=(5.00 kg)a49.1 N(11.00 kg)a=(5.00 kg)aa=49.1 N16.00 kg=3.07 m/s2

Substitute 3.07 m/s2 for a in equation IX and X to get the tension force.

    FT1=(3.00 kg)(3.07 m/s2)+29.4 N=38.6 N

    FT2=78.5 N(8.00 kg)(3.07 m/s2)=53.9 N

Therefore, the magnitude of tension on each rope is 38.6 N & 53.9 N and the acceleration on each block is 3.07 m/s2.

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