## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 3 - Vectors and Coordinates Systems - Exercises and Problems: 43

#### Answer

We should pull with 4.3 units of force at an angle of $83.4^{\circ}$ below the negative x-axis.

#### Work Step by Step

$F_x+3.0-(5.0)~sin(30^{\circ})= 0$ $F_x = -3.0+(5.0)~sin(30^{\circ})$ $F_x = -0.5~units~of~force$ $F_y + (5.0)~cos(30^{\circ}) = 0$ $F_y = -(5.0)~cos(30^{\circ})$ $F_y = -4.3~units~of~force$ We can find the magnitude of the applied force $F$; $F = \sqrt{(F_x)^2+(F_y)^2}$ $F = \sqrt{(-0.5)^2+(-4.3)^2}$ $F = 4.3~units~of~force$ We can find the angle $\theta$ below the negative x-axis; $tan(\theta) = \frac{4.3}{0.5}$ $\theta = arctan(\frac{4.3}{0.5})$ $\theta = 83.4^{\circ}$ We should pull with 4.3 units of force at an angle of $83.4^{\circ}$ below the negative x-axis.

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