Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Question: A bowler holds a bowling ball (M = 7.2 Kg) in the palm of his hand (Figure 12-37). His upper arm is vertical; his lower arm (1.8 kg) is horizontal. What is the magnitude of (a) the force of the biceps muscle on the lower arm and (b) the force between the bony structures at the elbow contact point?

Answer

The magnitude of the force of the biceps muscle on the lower arm, $T = 6.5 \times 10^{2} N$ .
Force between the bony structures at the elbow contact point, $F = 5.6 \times 10^{2} N$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Understanding the given information

Distance between the elbow point of contact and the lower arm CM, $D = 0.15 m, d = 0.04 m$ .
Distance between the elbow point of contact and the palm, L =0.33m .
Mass of the ball, M =7.2 kg .
Mass of the lower arm, m =1.8 kg .

Step 2: Concept and formula used in the given question

The ball is held in hands and balanced. It is neither moving linearly nor rotating about any pivot point. So, it is in static equilibrium condition. By selecting a pivot point and applying the static equilibrium conditions, you can write the torque equation in terms of force and distance. Solving this equation, you would get the unknown tension and force.

Step 3: (a) Calculation for the force of the biceps muscle on the lower arm

Using the figure given in the problem and condition of static equilibrium, we can write torque and force equations as,

$\sum T o r q u e = 0 (d \times T) + (0 \times F) - (D \times m g) - (L \times M g) = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) \sum F_{x} = 0 F + (M + m) g - T = 0 . . . . . . . . . . . . . . . . . . . . . . . . . (2)$

Substitute the values in equation 1, and we get,

$(0.04 \times T) + 0 - (0.15 \times 1.8 \times 9.8) - (0.33 \times 7.2 \times 9.8) = 0 T \times 0.04 = 25.9308 T = 648 N$

Hence, the magnitude of the force of the bicep muscle on the lower arm is $6.5 \times 10^{2} N$ .

Step 4: (b) Calculation for the force between the bony structures at the elbow contact point

Substitute the values in equation 2, and we get,

$F + (7.2 + 1.8) \times 9.8 - 648 = 0 F + 88.2 - 648 = 0 F \approx 560 N$

Hence, the force between the bony structures at the elbow contact point is $5.6 \times 10^{2} N$ .

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Understanding the given information

Step 2: Concept and formula used in the given question

Step 3: (a) Calculation for the force of the biceps muscle on the lower arm

Step 4: (b) Calculation for the force between the bony structures at the elbow contact point

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Understanding the given information

Step 2: Concept and formula used in the given question

Step 3: (a) Calculation for the force of the biceps muscle on the lower arm

Step 4: (b) Calculation for the force between the bony structures at the elbow contact point

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