Solution 4.5b
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(New page: Image:teori4.gif Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math> and <math>\tan \alpha =\frac{V}{H} </math> Here <math>V=-3\ \text{N}</math> and <math>H=4\ \text{N}</math>....)
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Revision as of 11:02, 23 March 2010
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 {{F}^{2}}={{4}^{2}}+{{\left( -3 \right)}^{2}}=25
giving
\displaystyle F=5\ \text{N}
Also
\displaystyle \tan \alpha =\frac{-3}{4}=B1
giving
\displaystyle \alpha =36\textrm{.}9{}^\circ
