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4. Forces and Vectors

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(New page: 4. Forces and Vectors Key Points <math>\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{al...)
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Revision as of 15:32, 18 March 2009

4. Forces and Vectors

Key Points



F=Fcosi+Fcos(90−)j=Fcosi+Fsinj





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.




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.



Solution


28cos30i−28sin30j


Note the negative sign in the second term.

Example 4.3



Solution


−50cos44i−50sin44j


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,

F=42+82=80=894 N (to 3sf) 


The angle, , is given by,

=tan−148=634 


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 20cos50i+20sin50j

18 N −18j

25 N −25cos20i−25sin20j

15 N −15cos30i+15sin30j


Resultant Force = 20cos50−25cos20−15cos30i+20sin50−18−25sin20+15sin30j=−23627i−3730j


The magnitude is given by:


236272+37302=239 N (to 3sf) 


The angle  can be found using tan.


tan=373023627=90

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