Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts an average force of 1350 N horizontally. Each of the second team’s members has an average mass of 73 kg and exerts an average force of 1365 N horizontally.
(a) What is magnitude of the acceleration of the two teams?
(b) What is the tension in the section of rope between the teams?

(a) The magnitude of the acceleration of the two teams is 0.106 m/s².

(b) The tension in the rope is \(12214.8\;{\rm{N}}\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Theory

Apply Newton’s second law of motion.

\({F_{{\rm{net}}}} = ma\)

\(\left( {{F_2} - {F_1}} \right) = \left( {{m_1} + {{\rm{m}}_2}} \right)a\) …….…………. (i)

Here, F_net is the net force, n is the number of members, m₁ is the average mass of the first team member, m₂ is the average mass of the second team member, F₁ is the force exerted by the first team members, F₂ is the force exerted by the second team members, and a is the acceleration of the two teams.

Step 2: Given data

An average mass of team 1 = 68 kg.
An average force exerted by team 1 =1350 N.
the second team’s members' average mass is= 73 kg.
Exerted average force by second-team =1365 N.

Step 3: (a) Determine the force of friction between the losing player’s feet and the grass.

Substitute 68 kg for m₁, 73 kg for m₂, 1350 N for F₁, and 1365 N for F₂ in equation (i), and we get,

\(\begin{array}{c}1365\;{\rm{N}} - 1350\;{\rm{N}} = \left( {68 + 73} \right)\;{\rm{kg}} \times a\\15\;{\rm{N}} = 141\;{\rm{kg}} \times a\\a = \frac{{15\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}{{141\;{\rm{kg}}}}\\a = 0.106\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\end{array}\)

Hence, the magnitude of the acceleration of the two teams is 0.106 m/s².

Step 4: (b) Determine the tension in the section of rope between the two teams

Apply Newton’s second law of motion,

\(T - n{F_1} = n{m_1}a\)

Here, T is the tension in the rope.

Substitute 68 kg for m₁, 9 for n, and 1350 N for F₁ in the above expression, and we get,

\(\begin{array}{c}T - 9 \times 1350\;{\rm{N}} = 9 \times 68\;{\rm{kg}} \times 0.106\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\T - 12150\;{\rm{N}} = 64.872\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}\\T = \left( {64.872 + 12150} \right)\;{\rm{N}}\\T = 12214.8\;{\rm{N}}\end{array}\)

Hence, the tension in the rope is\(12214.8\;{\rm{N}}\).

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Theory

Step 2: Given data

Step 3: (a) Determine the force of friction between the losing player’s feet and the grass.

Step 4: (b) Determine the tension in the section of rope between the two teams

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Theory

Step 2: Given data

Step 3: (a) Determine the force of friction between the losing player’s feet and the grass.

Step 4: (b) Determine the tension in the section of rope between the two teams

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