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Q15P

Expert-verifiedFound in: Page 346

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Question: Forces ** ${\overrightarrow{\mathbf{F}}}_{{\mathbf{1}}}{\mathbf{\text{,}}}{\overrightarrow{\mathbf{F}}}_{{\mathbf{2}}}{\mathbf{}}{\mathbf{\text{and}}}{\overrightarrow{\mathbf{F}}}_{{\mathbf{3}}}$** ** ** 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 **${\overrightarrow{\mathbf{F}}}_{h}{\mathbf{}}{\mathbf{\text{and}}}{\overrightarrow{\mathbf{F}}}_{v}$

Answer

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

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

** **

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:

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

Using the given figure in the problem and applying static equilibrium conditions: $\sum {\text{F}}_{\text{x}}=0\phantom{\rule{0ex}{0ex}}{\text{F}}_{\text{h}}-{\text{F}}_{3}=0\xb7\xb7\xb7\xb7\xb7\xb7\left(1\right)\phantom{\rule{0ex}{0ex}}\sum {\text{F}}_{\text{y}}=0\phantom{\rule{0ex}{0ex}}{\text{F}}_{\text{v}}-{\text{F}}_{1}-{\text{F}}_{2}=0\xb7\xb7\xb7\xb7\xb7\xb7\left(2\right)\phantom{\rule{0ex}{0ex}}\sum =0\phantom{\rule{0ex}{0ex}}\left({\text{F}}_{\text{v}}\times \text{d}\right)-\left({\text{F}}_{2}\times \text{b}\right)-\left({\text{F}}_{3}\times \text{a}\right)=0\xb7\xb7\xb7\xb7\xb7\xb7\left(3\right)\phantom{\rule{0ex}{0ex}}$

From equation (1):

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

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

From equation (2):

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

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

From equation (3):

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

Hence, d = 1.3 m

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