## Fundamentals of Physics Extended (10th Edition)

Published by Wiley

# Chapter 5 - Force and Motion-I - Problems: 39b

#### Answer

The Tension in the cord $= 3.7 \times 10^{-3}$ $N$

#### Work Step by Step

To find the tension in cord ($T$), we balance the forces in vertical direction. So, the forces in vertical direction are: i) Vertical component of tension (upwards) $= Tcos37^{\circ} = \frac{4}{5}T$ ii) Weight (downwards) $=mg$ Assuming upward direction positive, writing the force equation with proper sign convention gives: $\frac{4}{5}T-mg=ma$ Since the sphere is at rest, net acceleration is zero. So the above equation can becomes: $\frac{4}{5}T-mg=0$ $\frac{4}{5}T=mg$ Substituting known values in the equation and solving gives : $\frac{4}{5}T =[3\times 10^{-4} kg]\times [9.8$ $m/s^{2}]$ $T=3.675\times 10^{-3} N$ $T \approx 3.7 \times 10^{-3}$ $N$

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