Solution 9.2c
From Mechanics
(Difference between revisions)
(New page: The sum of the forces is zero as the body has constant velocity. This means the sum of the forces in all directions is zero. Resolving horisontally, <math>\begin{align} & \to :\quad 42+...) |
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Resolving horisontally, | Resolving horisontally, | ||
| - | |||
<math>\begin{align} | <math>\begin{align} | ||
| - | & \to :\quad 42+P-Q\cos {{45}^{\circ }}=0 \\ | + | & \to :\quad 42\cos {{30}^{\circ }}+P-Q\cos {{45}^{\circ }}=0 \\ |
& \\ | & \\ | ||
\end{align}</math> | \end{align}</math> | ||
Resolving vertically, | Resolving vertically, | ||
| - | |||
<math>\begin{align} | <math>\begin{align} | ||
| Line 17: | Line 15: | ||
& Q=\frac{25+42\sin {{30}^{\circ }}}{\cos {{45}^{\circ }}}=\frac{46}{0\textrm{.}707}=65\textrm{.}1 \ \text{N} \\ | & Q=\frac{25+42\sin {{30}^{\circ }}}{\cos {{45}^{\circ }}}=\frac{46}{0\textrm{.}707}=65\textrm{.}1 \ \text{N} \\ | ||
\end{align}</math> | \end{align}</math> | ||
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| + | Substituting this value for <math>Q</math> in the first equation gives | ||
| + | |||
| + | <math>P=Q\cos {{45}^{\circ }}-42\cos {{30}^{\circ }}=65\textrm{.}1\times 0\textrm{.}707-42\times 0\textrm{.}866=46-36\textrm{.}4=9\textrm{.}6\ \text{N}</math> | ||
Current revision
The sum of the forces is zero as the body has constant velocity. This means the sum of the forces in all directions is zero.
Resolving horisontally,
\displaystyle \begin{align} & \to :\quad 42\cos {{30}^{\circ }}+P-Q\cos {{45}^{\circ }}=0 \\ & \\ \end{align}
Resolving vertically,
\displaystyle \begin{align} & \uparrow :\quad 25+42\sin {{30}^{\circ }}-Q\cos {{45}^{\circ }}=0 \\ & \\ & Q=\frac{25+42\sin {{30}^{\circ }}}{\cos {{45}^{\circ }}}=\frac{46}{0\textrm{.}707}=65\textrm{.}1 \ \text{N} \\ \end{align}
Substituting this value for \displaystyle Q in the first equation gives
\displaystyle P=Q\cos {{45}^{\circ }}-42\cos {{30}^{\circ }}=65\textrm{.}1\times 0\textrm{.}707-42\times 0\textrm{.}866=46-36\textrm{.}4=9\textrm{.}6\ \text{N}
