Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 4 - Forces and Newton's Laws of Motion - Problems - Page 115: 49

Answer

The minimum force needed to hold the picture static against the wall is $16.33$ $N$

Work Step by Step

Let $ F$ be the applied force on the picture and $m$ be the mass of the picture. Now the forces acting on $m$ are (a) weight ($mg$) of the picture in downward direction, (b) friction ($f$) in upward direction, (c) normal force ($N$) in outward direction from the wall, (d) external force ($ F$) towards the wall As the picture is in static condition, total resultant forces acting on the picture would be zero. Thus, $f=mg$ and $N=F$ Again, we know that $f=\mu N$, where $\mu$ is the coefficient of static friction between the picture and the wall. Thus, $mg=\mu F$ or, $F=\frac{mg}{\mu}$ Putting $m=1.10$ $kg$, $g=9.8$ $m/s^{2}$, and $\mu =0.660$, we get, $f=16.33$ $N$ Therefore the minimum force needed to hold the picture static against the wall is $16.33$ $N$
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