Solution 5.3b

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(New page: From the previous part we have that <math>\mathbf{F}4=-\mathbf{F}1-\mathbf{F}2-\mathbf{F}3=-292\mathbf{i}-62\mathbf{j}\ \text{N}</math> Thus the magnitude <math>Q</math> is given by <ma...)
Current revision (12:12, 4 April 2010) (edit) (undo)
 
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<math>Q{}^{2}={{\left( -292 \right)}^{2}}+{{\left( -62 \right)}^{2}}=85300+3840=89140</math>
<math>Q{}^{2}={{\left( -292 \right)}^{2}}+{{\left( -62 \right)}^{2}}=85300+3840=89140</math>
- 
or
or
<math>Q=299 \text{ N}</math>
<math>Q=299 \text{ N}</math>
 +
 +
and
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 +
<math>\tan \alpha =\frac{62\textrm{.}9}{292}=0\textrm{.}215</math>
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 +
giving
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<math>\alpha ={{12\textrm{.}2}^{\circ }}</math>

Current revision

From the previous part we have that

\displaystyle \mathbf{F}4=-\mathbf{F}1-\mathbf{F}2-\mathbf{F}3=-292\mathbf{i}-62\mathbf{j}\ \text{N}

Thus the magnitude \displaystyle Q is given by

\displaystyle Q{}^{2}={{\left( -292 \right)}^{2}}+{{\left( -62 \right)}^{2}}=85300+3840=89140

or

\displaystyle Q=299 \text{ N}

and

\displaystyle \tan \alpha =\frac{62\textrm{.}9}{292}=0\textrm{.}215

giving

\displaystyle \alpha ={{12\textrm{.}2}^{\circ }}

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