Solution 4.5a

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Current revision (10:42, 23 March 2010) (edit) (undo)
 
Line 9: Line 9:
Here
Here
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<math>V=A</math> and
+
<math>V=3\ \text{N}</math> and
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<math>H=B</math>.
+
<math>H=4\ \text{N}</math>.
Substituting the given values
Substituting the given values
- 
<math>\begin{align}
<math>\begin{align}
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& {{F}^{2}}={{4}^{2}}+{{3}^{2}}=25 \\
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{{F}^{2}}={{4}^{2}}+{{3}^{2}}=25 \\
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& \text{giving} \\
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& F=5\ \text{N} \\
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\end{align}</math>
\end{align}</math>
 +
giving
 +
 +
<math>\begin{align}
 +
F=5\ \text{N} \\
 +
\end{align}</math>
Also
Also

Current revision

Image:teori4.gif

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

and

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

Here

\displaystyle V=3\ \text{N} and \displaystyle H=4\ \text{N}.

Substituting the given values

\displaystyle \begin{align} {{F}^{2}}={{4}^{2}}+{{3}^{2}}=25 \\ \end{align}

giving

\displaystyle \begin{align} F=5\ \text{N} \\ \end{align}

Also

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

giving

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