Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 6 - Applications of Newton's Laws - Problems and Conceptual Exercises - Page 184: 83

Answer

(a) greater than (b) $3.4N$ (c) $0.41Kg$

Work Step by Step

(a) We know that the horizontal component of string 1 balances the large horizontal component of string 2. Thus, string 1 supports most of the weight of the picture and hence the tension in string 1 is greater than string 2. (b) We can calculate the tension in string 1 as $\Sigma F_x=-T_1cos\theta+T_2cos\theta=0$ This simplifies to: $T_1=T_2 \frac{cos\theta_2}{cos\theta_1}$ We plug in the known values to obtain: $T_1=(1.7)\frac{cos 32^{\circ}}{cos 65^{\circ}}$ $T_1=3.4N$ (c) We can find the mass of the picture as follows: $\Sigma F_y=T_1sin\theta_1+T_2sin\theta_2-W=0$ $\implies W=T_1sin\theta+T_2sin\theta_2$ We plug in the known values to obtain: $W=(3.4)sin65^{\circ}+(1.7)sin32^{\circ}$ $W=4.0N$ Now $m=\frac{W}{g}$ We plug in the known values to obtain: $m=\frac{4.0}{9.81}$ $m=0.41Kg$
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