Answer
$0.85N$
Work Step by Step
We know that
$\Sigma F_y=Ncos30^{\circ}+Ncos30^{\circ}-mg=0$
This can be rearranged as:
$N=\frac{mg}{2cos30^{\circ}}$
We plug in the known values to obtain:
$N=\frac{(0.15)(9.81)}{2cos30^{\circ}}$
$N=0.85N$
Physics Technology Update (4th Edition)
by Walker, James S.
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 181: 39
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