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Found in: Page 346

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

Question: Forces ${\stackrel{\mathbf{\to }}{\mathbf{F}}}_{{\mathbf{1}}}{\mathbf{\text{,}}}{\stackrel{\mathbf{\to }}{\mathbf{F}}}_{{\mathbf{2}}}{\mathbf{}}{\mathbf{\text{and}}}{\stackrel{\mathbf{\to }}{\mathbf{F}}}_{{\mathbf{3}}}$ act on the structure of Fig. 12-33, shown in an overhead view. We wish to put the structure in equilibrium by applying a fourth force, at a point such as P. The fourth force has vector components ${\stackrel{\mathbf{\to }}{\mathbf{F}}}_{h}{\mathbf{}}{\mathbf{\text{and}}}{\stackrel{\mathbf{\to }}{\mathbf{F}}}_{v}$ . We are given that a = 2.0 m, b = 3.0m , c = 1 0 m , ${\stackrel{\mathbf{\to }}{\mathbf{F}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{20}}{\mathbf{}}{\mathbf{\text{N ,}}}{\stackrel{\mathbf{\to }}{\mathbf{F}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{10}}{\mathbf{}}{\mathbf{\text{N and}}}{\stackrel{\mathbf{\to }}{\mathbf{F}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{5}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\text{N}}}$Find (a) Fh , (b) Fv , and (c) d.

$a.{F}_{h}=5.0\text{N}\phantom{\rule{0ex}{0ex}}b.{F}_{v}=30\text{N}\phantom{\rule{0ex}{0ex}}c.d=1.3m$

See the step by step solution

Step 1: Understanding the given information

$\text{a}=2.0\text{m}\phantom{\rule{0ex}{0ex}}\text{b}=3.0\text{m}\phantom{\rule{0ex}{0ex}}\text{c}=1.0\text{m}\phantom{\rule{0ex}{0ex}}{\text{F}}_{1}=20\text{N}\phantom{\rule{0ex}{0ex}}{\text{F}}_{2}=10\text{N}\phantom{\rule{0ex}{0ex}}{F}_{3}=5.0\text{N}\phantom{\rule{0ex}{0ex}}$

Step 2: Concept and formula used in the given question

By applying the equations of static equilibrium, you can get the equations in terms of unknown forces. By solving these equations, you can find the values of unknown forces and distance. The equations used are given below.

Static Equilibrium conditions:

$\sum {F}_{x}=0\phantom{\rule{0ex}{0ex}}\sum {F}_{y}=0\phantom{\rule{0ex}{0ex}}\sum \tau =0\phantom{\rule{0ex}{0ex}}$

Step 3: (a) Calculation for the Fh

Using the given figure in the problem and applying static equilibrium conditions: $\sum {\text{F}}_{\text{x}}=0\phantom{\rule{0ex}{0ex}}{\text{F}}_{\text{h}}-{\text{F}}_{3}=0······\left(1\right)\phantom{\rule{0ex}{0ex}}\sum {\text{F}}_{\text{y}}=0\phantom{\rule{0ex}{0ex}}{\text{F}}_{\text{v}}-{\text{F}}_{1}-{\text{F}}_{2}=0······\left(2\right)\phantom{\rule{0ex}{0ex}}\sum =0\phantom{\rule{0ex}{0ex}}\left({\text{F}}_{\text{v}}×\text{d}\right)-\left({\text{F}}_{2}×\text{b}\right)-\left({\text{F}}_{3}×\text{a}\right)=0······\left(3\right)\phantom{\rule{0ex}{0ex}}$

From equation (1):

${F}_{h}-5=0$

Hence, ${F}_{h}-5=0$

Step 4: (b) Calculation for the  Fv

From equation (2):

${F}_{v}-20-10=0$

Hence, ${F}_{v}=30N$

Step 5: (c) Calculation for the  d

From equation (3):

$\left(30×d\right)-\left(10×3\right)-\left(5×2\right)=0\phantom{\rule{0ex}{0ex}}d=1.3\text{m}\phantom{\rule{0ex}{0ex}}$

Hence, d = 1.3 m