Solution 4.5a

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(New page: Image:teori4.gif Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math> Substituting the given values <math>\begin{align} & {{F}^{2}}={{4}^{2}}+{ {3}^{2}} =25 \\ & F=5 \text{ N} \\ \end...)
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Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math>
Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math>
 +
 +
and
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<math>\tan \alpha =\frac{V}{H} </math>
 +
 +
Here
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 +
<math>V=A</math> and
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<math>H=B</math>.
Substituting the given values
Substituting the given values
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<math>\begin{align}
<math>\begin{align}
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& {{F}^{2}}={{4}^{2}}+{
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& {{F}^{2}}={{4}^{2}}+{{3}^{2}}=25 \\
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{3}^{2}}
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& \text{giving} \\
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=25 \\
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& F=5\ \text{N} \\
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& F=5 \text{ N} \\
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\end{align}</math>
\end{align}</math>
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<math>\tan \alpha =\frac{V}{H}</math>
 
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Here
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Also
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<math>V=3\ \text{N}</math>
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-
and
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<math>H=4\ \text{N}</math>.
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-
 
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Thus
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<math>\tan \alpha =\frac{3}{4}=0\textrm{.}75</math>
<math>\tan \alpha =\frac{3}{4}=0\textrm{.}75</math>
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giving
giving
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<math>\alpha =36\textrm{.}9{}^\circ</math>
<math>\alpha =36\textrm{.}9{}^\circ</math>

Revision as of 10:35, 23 March 2010

Image:teori4.gif

Using \displaystyle {{F}^{2}}={{H}^{2}}+{{V}^{2}}

and

\displaystyle \tan \alpha =\frac{V}{H}

Here

\displaystyle V=A and \displaystyle H=B.

Substituting the given values


\displaystyle \begin{align} & {{F}^{2}}={{4}^{2}}+{{3}^{2}}=25 \\ & \text{giving} \\ & F=5\ \text{N} \\ \end{align}


Also

\displaystyle \tan \alpha =\frac{3}{4}=0\textrm{.}75

giving

\displaystyle \alpha =36\textrm{.}9{}^\circ