Processing Math: 53%
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.

No jsMath TeX fonts found -- using image fonts instead.
These may be slow and might not print well.
Use the jsMath control panel to get additional information.
jsMath Control PanelHide this Message


jsMath

4. Forces and Vectors

From Mechanics

Jump to: navigation, search

F=Fcosi+Fcos(90)j=Fcosi+Fsinj


Image:TF.teori.GIF



Fcos is one component of the force. If i is horizontal, Fcos is called the horizontal component of the force.


Fsin is another component of the force. If j is vertical, Fsin is called the vertical component of the force.

Example 4.1

Express each of the forces given below in the form ai + bj.

(a)

Image:TF4.1a.GIF

(b)

Image:TF4.1b.GIF

Solution

(a)

20cos40i+20sin40j


(b)

80cos30i+80sin30j


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.


Image:TF4.2.GIF


Solution


28cos30i28sin30j


Note the negative sign in the second term.


Example 4.3


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


Image:TF4.3.GIF


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

Solution

Image:TF4.4.GIF


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, \displaystyle \theta , 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.


Image:TF4.5.GIF


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 \displaystyle \theta can be found using tan.


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


Image:TF4.5a.GIF