Förberedande Mekanik

Versionen från 18 mars 2009 kl. 15.32 (redigera) Ian (Diskussion | bidrag) (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...) ← Gå till föregående ändring		Nuvarande version (19 mars 2009 kl. 15.01) (redigera) (ogör) Ian (Diskussion | bidrag)
(2 mellanliggande versioner visas inte.)
Rad 1:		Rad 1:
-	4. Forces and Vectors
-
-	Key Points
-
-
-
-
	<math>\begin{align}		<math>\begin{align}
	& \mathbf{F}=F\cos \alpha \mathbf{i}+F\cos (90-\alpha )\mathbf{j} \\		& \mathbf{F}=F\cos \alpha \mathbf{i}+F\cos (90-\alpha )\mathbf{j} \\
Rad 13:		Rad 6:
-		+	[[Image:TF.teori.GIF]]
Rad 20:		Rad 13:
	<math>F\cos \alpha </math>		<math>F\cos \alpha </math>
-	is one component of the force. If i is horizontal,	+	is one component of the force. If <math>\mathbf{i}</math> is horizontal,
	<math>F\cos \alpha </math>		<math>F\cos \alpha </math>
	is called the horizontal component of the force.		is called the horizontal component of the force.
Rad 26:		Rad 19:
	<math>F\sin \alpha </math>		<math>F\sin \alpha </math>
-	is another component of the force. If j is vertical,	+	is another component of the force. If <math>\mathbf{j}</math> is vertical,
	<math>F\sin \alpha </math>		<math>F\sin \alpha </math>
	is called the vertical component of the force.		is called the vertical component of the force.
		+	'''[[Example 4.1]]'''
		+	Express each of the forces given below in the form a<math>\mathbf{i}</math> + b<math>\mathbf{j}</math>.
-	Example 4.1	+	(a)
-	Express each of the forces given below in the form ai + bj.	+	[[Image:TF4.1a.GIF]]
		+	(b)
		+	[[Image:TF4.1b.GIF]]
-		+	'''Solution'''
-		+
-		+
-	Solution	+
	(a)		(a)
Rad 54:		Rad 48:
	Note the negative sign here in the first term.		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.
		+
		+	'''[[Example 4.2]]'''
		+
		+	Express the force shown below as a vector in terms of <math>\mathbf{i}</math> and <math>\mathbf{j}</math>.
		+	[[Image:TF4.2.GIF]]
-	Solution	+	'''Solution'''
Rad 71:		Rad 67:
	Note the negative sign in the second term.		Note the negative sign in the second term.
-	Example 4.3
		+	'''[[Example 4.3]]'''
		+	Express the force shown below as a vector in terms of
		+	<math>\mathbf{i}</math>
		+	and
		+	<math>\mathbf{j}</math>
-	Solution	+
		+	[[Image:TF4.3.GIF]]
		+
		+
		+	'''Solution'''
Rad 85:		Rad 89:
	Note that here both terms are negative.		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	+	'''[[Example 4.4]]'''
		+
		+	Find the magnitude of the force (4<math>\mathbf{i}</math> - 8<math>\mathbf{j}</math>) 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,		The magnitude, FN , of the force is given by,
Rad 96:		Rad 105:
-	The angle, , is given by,	+	The angle, <math>\theta </math>, is given by,
	<math>\theta ={{\tan }^{-1}}\left( \frac{8}{4} \right)=63.4{}^\circ </math>		<math>\theta ={{\tan }^{-1}}\left( \frac{8}{4} \right)=63.4{}^\circ </math>
-	Example 4.5	+	'''[[Example 4.5]]'''
	Find the magnitude and direction of the resultant of the four forces shown in the diagram.		Find the magnitude and direction of the resultant of the four forces shown in the diagram.
		+	[[Image:TF4.5.GIF]]
-	Solution	+	'''Solution'''
	Force Vector Form		Force Vector Form
Rad 138:		Rad 148:
-	The angle  can be found using tan.	+	The angle <math>\theta </math> can be found using tan.
Rad 145:		Rad 155:
	& \theta =9.0{}^\circ		& \theta =9.0{}^\circ
	\end{align}</math>		\end{align}</math>
		+
		+
		+	[[Image:TF4.5a.GIF]]

---

Nuvarande version

\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}

(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 \displaystyle \mathbf{i} and \displaystyle \mathbf{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

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}

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, 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.

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}