Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 9 - Static Equilibrium; Elasticity and Fracture - Problems: 11

Answer

A) The pivot should be 2.3m from the adult. B) The pivot should be 2.5m from the adult.

Work Step by Step

A) Let x be the distance from the child to the pivot point. Let y be the distance from the adult to the pivot point. Because the board is 9m long, we know that: $x+y=9m$ $x=9m-y$ Now we will sum the torques about the pivot point. $(+ \circlearrowleft) \sum \tau_p =0$ $x*g*25kg-75kg*g*y=0$ $(9m-y)g*25kg-75kg*g*y=0$ $9m*g*25kg-y*g*25kg-75kg*g*y=0$ $y=\frac{9m*g*25kg}{g*25kg+g*75kg}$ $y=\frac{9m*9.8m/s^2*25kg}{9.8m/s^2*25kg+9.8m/s^2*75kg}$ $y\approx2.3m$ The pivot should be 2.3m from the adult. B) We will use the same approach but with the addition of the boards weight. Because the board is 9m long, we know that: $x+y=9m$ $x=9m-y$ Now we will sum the torques about the pivot point. $(+ \circlearrowleft) \sum \tau_p =0$ $x*g*25kg+(4.5m-x)*g*15kg-75kg*g*y=0$ $(9m-y)g*25kg+(4.5m-(9m-y))*g*15kg-75kg*g*y=0$ $(9m-y)g*25kg+(4.5m-9m+y)*g*15kg-75kg*g*y=0$ $9m*g*25kg+(-4.5m+y)*g*15kg-y*g*25kg-75kg*g*y=0$ $9m*g*25kg-4.5m*g*15kg+y*g*15kg-y*g*25kg-75kg*g*y=0$ $9m*g*25kg-4.5m*g*15kg+y(g*15kg-g*25kg-75kg*g)=0$ $y=\frac{-9m*g*25kg+4.5m*g*15kg}{(g*15kg-g*25kg-75kg*g)}$ $y=\frac{-9m*25kg+4.5m*15kg}{(15kg-25kg-75kg)}$ $y\approx2.5m$ The pivot should be 2.5m from the adult.
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