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 - Page 252: 9


The reaction force near the piano is 2890N. The reaction away from the piano is 1320N

Work Step by Step

We will sum the torques about the support near the piano and then sum the force vertically to find the reactions in the supports. Let $F_C$ be the reaction force near the piano and $F_A$ be the other reaction. $(+ \circlearrowleft) \sum \tau_c =0$ $-320kg*9.8m/s^2*1m-110kg*9.8m/s^2*2m+F_R*4m=0$ $F_R=\frac{320kg*9.8m/s^2*1m+110kg*9.8m/s^2*2m}{4m}$ $F_R\approx1320N$ $(\uparrow +) \sum \overrightarrow{F}_{y} =0 $ $F_C+F_R-320kg*9.8m/s^2-110kg*9.8m/s^2=0$ $F_C=-F_R+320kg*9.8m/s^2+110kg*9.8m/s^2=0$ $F_C\approx2890N$
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