College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 3 - Problems - Page 115: 67


$a = 2.08~m/s^2$

Work Step by Step

Let $F_T$ be the tension in the pendulum. Let $m$ be the mass of the bob. The vertical component of $F_T$ is equal in magnitude to the bob's weight. $F_T~cos~\theta = mg$ $F_T = \frac{mg}{cos~\theta}$ We can find the acceleration: $ma = F_T~sin~\theta$ $ma = \frac{mg~sin~\theta}{cos~\theta}$ $a = g~tan~\theta$ $a = (9.80~m/s^2)~tan~12^{\circ}$ $a = 2.08~m/s^2$
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