Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

In Fig. 12-69, a package of mass m hangs from a short cord that is tied to the wall via cord 1 and to the ceiling via cord 2. Cord 1 is at angle $ϕ = 40 °$ with the horizontal; cord 2 is at angle $θ$ . (a) For what value of $θ$ is the tension in cord 2 minimized? (b) In terms of mg, what is the minimum tension in cord 2

a) The angle of cord 2 with the horizontal i.e. $θ = 50 °$ .

b) The minimum tension in cord 2 in terms of mg is the $0.77mg$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Listing the given quantities

$ϕ = 40 °$

Step 2: Understanding the concept of force and tension

From the Free Body Diagram and condition for static equilibrium of the system, we can find the angle and the tension in cord 2.

Equations:

$\sum_{}^{} F_{x} = 0$

$\sum_{}^{} F_{y} = 0$

$Weight = mg$

Step 3: Free body diagram

Step 4: (a) Calculations of the angle of cord 2 with the horizontal

$\begin{array}{l} \sum_{}^{} F_{x} = 0 \\ T_{1} cos 40 ° - T_{2} cos θ = 0 \\ T_{1} cos 40 ° = T_{2} cos θ \\ T_{1} = \frac{T_{2} cos θ}{cos 40 °} \end{array}$

$\begin{array}{l} \sum_{}^{} F_{y} = 0 \\ T_{1} {sin 40° +T}_{2} sinθ - mg = 0 \\ \frac{T_{2} cos θ}{cos 40 °} \times {sin 40°+T}_{2} sinθ = mg \\ T_{2} (sinθ + cos θtan40 °) = mg \\ T_{2} = \frac{mg}{(sinθ + cosθ tan40 °)} \end{array}$

As the value in the $T_{2}$ should be minimum, we take derivative at both sides, and we get,

$\begin{array}{l} \frac{d}{d θ} (T_{2}) = \frac{d}{d θ} [\frac{mg}{(sinθ + cosθ tan40 °)}] \\ 0 = \frac{- mg (tan 40 ° \times - sin θ + cos θ)}{{((tan 40 ° ×cos θ) +sinθ)}^{2}} \\ (tan 40 ° \times - sin θ + cos θ) = 0 \\ tan 40 ° = \cot θ \\ θ = \cot^{- 1} (tan 40 °) \\ θ = 50 ° \end{array}$

Step 5: (b) Calculations of the minimum tension in the cord in terms of mg

As we know,

$\begin{matrix} T_{2} = \frac{mg}{(tan 40 ° ×cos θ) +sinθ} \\ = \frac{mg}{(tan 40 ° ×cos 50 °) +sin 50 °} \\ =0.77mg \end{matrix}$

The minimum tension in cord 2 in terms of mg is the $0.77mg$ .

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Listing the given quantities

Step 2: Understanding the concept of force and tension

Step 3: Free body diagram

Step 4: (a) Calculations of the angle of cord 2 with the horizontal

Step 5: (b) Calculations of the minimum tension in the cord in terms of mg

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Fundamentals Of Physics

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Step by Step Solution

Step 1: Listing the given quantities

Step 2: Understanding the concept of force and tension

Step 3: Free body diagram

Step 4: (a) Calculations of the angle of cord 2 with the horizontal

Step 5: (b) Calculations of the minimum tension in the cord in terms of mg

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