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	== '''Key Points''' ==		== '''Key Points''' ==
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		+	The force <math>\mathbf{F}</math> can be resolved into components as follows.
	<math>\begin{align}		<math>\begin{align}

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Revision as of 13:31, 8 March 2010

Theory

Exercises

Key Points

The force \displaystyle \mathbf{F} can be resolved into components as follows.

\displaystyle \begin{align} & \mathbf{F}=F\cos \alpha \mathbf{i}+F\cos (90-\alpha )\mathbf{j} \\ & =F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}

\displaystyle F\cos \alpha is one component of the force. If \displaystyle \mathbf{i} is horizontal, \displaystyle F\cos \alpha is called the horizontal component of the force.

\displaystyle F\sin \alpha is another component of the force. If \displaystyle \mathbf{j} is vertical, \displaystyle F\sin \alpha is called the vertical component of the force.

Example 4.1

Express each of the forces given below in the form a\displaystyle \mathbf{i} + b\displaystyle \mathbf{j}.

a)

b)

Solution

a) \displaystyle 20\cos 40{}^\circ \mathbf{i}+20\sin 40{}^\circ \mathbf{j}\ \text{N}

b) \displaystyle -80\cos 30{}^\circ \mathbf{i}+80\sin 30{}^\circ \mathbf{j}\ \text{N}

Note the negative sign here in the first term.

Example 4.2

Express the force shown below as a vector in terms of \displaystyle \mathbf{i} and \displaystyle \mathbf{j}.

Solution

\displaystyle 28\cos 30{}^\circ \mathbf{i}-28\sin 30{}^\circ \mathbf{j}\ \text{N}

Note the negative sign in the second term.

Example 4.3

Express the force shown below as a vector in terms of \displaystyle \mathbf{i} and \displaystyle \mathbf{j}

Solution

\displaystyle -50\cos 44{}^\circ \mathbf{i}-50\sin 44{}^\circ \mathbf{j}\ \text{N}

Note that here both terms are negative.

Example 4.4

Find the magnitude of the force (4\displaystyle \mathbf{i} - 8\displaystyle \mathbf{j}) N. Draw a diagram to show the direction of this force.

Solution

The magnitude, \displaystyle F , of the force is given by,

\displaystyle F=\sqrt{{{4}^{2}}+{{8}^{2}}}=\sqrt{80}=8\textrm{.}94\text{ N (to 3sf)}

The angle, \displaystyle \theta , is given by,

\displaystyle \theta ={{\tan }^{-1}}\left( \frac{8}{4} \right)=63\textrm{.}4{}^\circ

Example 4.5

Find the magnitude and direction of the resultant of the four forces shown in the diagram.

Solution

Force	Vector Form (N)
20 N	\displaystyle 20\cos 50{}^\circ \mathbf{i}+20\sin 50{}^\circ \mathbf{j}
18 N	\displaystyle -18\mathbf{j}
25 N	\displaystyle -25\cos 20{}^\circ \mathbf{i}-25\sin 20{}^\circ \mathbf{j}
15 N	\displaystyle -15\cos 30{}^\circ \mathbf{i}+15\sin 30{}^\circ \mathbf{j}

\displaystyle \begin{align} & \text{Resultant Force }=\text{ }\left( 20\cos 50{}^\circ -25\cos 20{}^\circ -15\cos 30{}^\circ \right)\mathbf{i}+\left( 20\sin 50{}^\circ -18-25\sin 20{}^\circ +15\sin 30{}^\circ \right)\mathbf{j} \\ & =-23\textrm{.}627\mathbf{i}-3\textrm{.}730\mathbf{j} \ \text{N} \end{align}

The magnitude is given by:

\displaystyle \sqrt{{{23\textrm{.}627}^{2}}+{{3\textrm{.}730}^{2}}}=23\textrm{.}9\text{ N (to 3sf)}

The angle \displaystyle \theta can be found using tan.

\displaystyle \begin{align} & \tan \theta =\frac{3\textrm{.}730}{23\textrm{.}627} \\ & \theta =9\textrm{.}0{}^\circ \end{align}