Processing Math: Done

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Revision as of 14:50, 19 March 2009 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (15:01, 19 March 2009) (edit) (undo) Ian (Talk | contribs)
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	'''Solution'''		'''Solution'''
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	The magnitude, FN , of the force is given by,		The magnitude, FN , of the force is given by,
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	& \theta =9.0{}^\circ		& \theta =9.0{}^\circ
	\end{align}</math>		\end{align}</math>
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Current revision

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.

(a)

(b)

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

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

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