Suggested languages for you:

Americas

Europe

Q29 P

Expert-verified
Found in: Page 73

### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# The helicopter view in Fig. P3.29 (page 74) shows two W people pulling on a stubborn mule. The person on the right pulls with a force ${\stackrel{\to }{F}}_{1}$ of magnitude $120N$ and direction of ${\theta }_{1}=60.0°$ .The person on the left pulls with a force $80.0N$ of magnitude ${\theta }_{2}=75.0°$ and direction of . Find (a) the single force that is equivalent to the two forces shown and (b) the force that a third person would have to exert on the mule to make the resultant force equal to zero. The forces are measured in units of newtons (symbolized N).

$a\right)|\stackrel{\to }{F}|=185Nat\theta =77.8°\phantom{\rule{0ex}{0ex}}b\right)\left(-39.3\stackrel{^}{i}-181\stackrel{^}{j}\right)$

See the step by step solution

## Step 1: Define magnitude

• In simple terms, magnitude means' distance or number.' It describes the absolute or relative size or direction in which an object moves in the feeling of motion.
• It's a term for describing the size or scope of something. In general, magnitude in physics refers to distance or amount.

## Step 2 (a): Determine the magnitude and angle of the single force.

We must first find the resultant force in the x and y directions before we can find the single force.

${F}_{x}={F}_{1}cos{\theta }_{1}-{F}_{2}cos{\theta }_{2}\phantom{\rule{0ex}{0ex}}=120×cos60°-80×cos75°\phantom{\rule{0ex}{0ex}}=39.3N\phantom{\rule{0ex}{0ex}}{F}_{y}={F}_{1}sin{\theta }_{1}+{F}_{2}sin{\theta }_{2}\phantom{\rule{0ex}{0ex}}=120×sin60°+80×sin75°\phantom{\rule{0ex}{0ex}}=181N$

As a result, the magnitude is determined by:

$|\stackrel{\to }{F}|=\sqrt{{F}_{x}^{2}+{F}_{y}^{2}}\phantom{\rule{0ex}{0ex}}|\stackrel{\to }{F}|=\sqrt{39.{3}^{2}+{181}^{2}}\phantom{\rule{0ex}{0ex}}|\stackrel{\to }{F}|=185N$

In addition, the angle is determined by:

$\theta =ta{n}^{-1}\frac{y}{x}\phantom{\rule{0ex}{0ex}}=ta{n}^{-1}\frac{181}{39.3}\phantom{\rule{0ex}{0ex}}=77.8°\phantom{\rule{0ex}{0ex}}$

Hence, the single force is $|\stackrel{\to }{F}|=185Nat\theta =77.8°$

## Step 3 (b): Determine the magnitude and angle in the opposite direction.

${F}_{x}=39.3N\phantom{\rule{0ex}{0ex}}{F}_{y}=181N$

Hence, the magnitude and angle in the opposite direction is $\left(-39.3\stackrel{^}{i}-181\stackrel{^}{j}\right)N$ .