From Mechanics

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Revision as of 15:32, 18 March 2009

4. Forces and Vectors

Key Points

\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 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 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 ai + bj.

Solution

(a)

\displaystyle 20\cos 40{}^\circ \mathbf{i}+20\sin 40{}^\circ \mathbf{j}

(b)

\displaystyle -80\cos 30{}^\circ \mathbf{i}+80\sin 30{}^\circ \mathbf{j}

Note the negative sign here in the first term.

Example 4.2

Express the force shown below as a vector in terms of i and j.

Solution

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

Note the negative sign in the second term.

Example 4.3

Solution

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

Note that here both terms are negative.

Example 4.4

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

Solution

The magnitude, FN , of the force is given by,

\displaystyle F=\sqrt{{{4}^{2}}+{{8}^{2}}}=\sqrt{80}=8.94\text{ N (to 3sf)}

The angle, , is given by,

\displaystyle \theta ={{\tan }^{-1}}\left( \frac{8}{4} \right)=63.4{}^\circ

Example 4.5

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

Solution

Force Vector Form 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} \\ & =-23.627\mathbf{i}-3.730\mathbf{j} \end{align}

The magnitude is given by:

\displaystyle \sqrt{{{23.627}^{2}}+{{3.730}^{2}}}=23.9\text{ N (to 3sf)}

The angle  can be found using tan.

\displaystyle \begin{align} & \tan \theta =\frac{3.730}{23.627} \\ & \theta =9.0{}^\circ \end{align}