How do I solve problem 3.9b in my homework?

[khizman]

http://img362.imageshack.us/img362/8182/untitled128sh.jpg
how on Earth are you meant to do 3.9
i have tried using pythagoras's theorem but it doesn't work as i have the answer on the back of my book and it doesn't match up,
any ideas?

 

[Astronuc]

Since the ship is traveling along XO, the resultant force must be along this line, and to maintain this orientation, the lateral force components of forces F_A and F_B must be equal.

The magnitude of th lateral component is just F * sin \alpha where \alpha is the angle between XO and the force, so for F_B, \alpha = \theta.

The resultant is then just the sum of the vertical force components of forces F_A and F_B.

 

[khizman]

Since the ship is traveling along XO, the resultant force must be along this line, and to maintain this orientation, the lateral force components of forces F_A and F_B must be equal.

The magnitude of th lateral component is just F * sin \alpha where \alpha is the angle between XO and the force, so for F_B, \alpha = \theta.

The resultant is then just the sum of the vertical force components of forces F_A and F_B.

I really don't understand

 

[big man]

He is saying that there is a vertical and horizontal component to each of those force vectors. Since the ship is traveling along the line from XO the 'vertical' components of the TugA vector and TugB vector will combine to form the resultant force since the 'horizontal' components effectively cancel. The 'horizontal' components cancel because they are equal in magnitude (as Astronuc said), but opposite in direction.

F_R = F_A_y + F_B_y

Then if you use your rules from right-angle trigonometry you will get that the vertical component of TugB for example is:

F_B_y = F_B*cos\theta

 

[Astronuc]

A vector has both magnitude and direction.

In 2_D, a vector can be resolved into two normal components, e.g. in Cartesian coordinates, x and y.

In your problem, think of x as lateral (+x to right, -x to left) and y as vertical.

\vec{F_A} can be resolved as \vec{F_Ax} + \vec{F_Ay}.

Let F_A be the magnitude of \vec{F_A}, then the magnitude of \vec{F_Ax} is just F_Ax = F_A*sin 15°, and F_Ay = F_A*cos 15°

also F_A = \sqrt{(F_Ax)^2 + (F_Ay)^2)} - as in the Pythogorean relationship.

Now for the ship to travel in a straight line XO, the net lateral force must equal zero, so the lateral component of \vec{F_Ax} = -\vec{F_Bx}, and the magnitude of this vector, F_Bx is just F_B * sin \theta.

 

[iruka]

I am not a teacher and I am not experienced at explaining or solving mathematics problems.
I reply to khizmans post in order to test my understanding.


For problem 3.9b I can make a few suggestions.

algebraic methods are easily used with x,y component form of vectors

**hints to determine angle theta of Tug B vector**

>>Define x-axis and y-axis
I recommend using point x as the origin.

>>Take a look at the angles of vectors in the problem:

resultant is... 90 degrees to positive x-axis. [cos 90(resultant kN)]
Tug A is...(90 +15) degrees to the positive x-axis.
Tug B is (90 - theta) degrees to the positive x-axis.

Take note of the resultants length along the x-axis

>>The x-axis is cosine of the angle.

Some transposition of terms will give you Tug B theta.


**hints to determine the magnitude of the resultant vector**

>>The y-axis is the sine of the angle.

Maybe mapping the vectors on a graph would be easier than the algebraic method.


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A lot of explanation that may help with solving the problem 3.9b

If you spot an error let me know .
Some suggestions on better formatting are welcome.


The first problem to be solved is stated as:

Two tugs A and B pull a ship along the direction XO. See figure.
Tug A exerts a force on the ship of 3.0*10^4 N a an angle of 15 degrees to XO.
Tub B pulls with a force of 1.8*10^4 at an angle of theta to XO.
i Find the angle theta for the resultant force on the boat along XO.
ii Find the value of resultant force.


Before anything else can be done we need to be aware of what a vector is.
a vector is a magnitude with a direction. we are given force magnitudes and
the angle of forces from a reference direction.
( magnitude is shown graphically as a length.)

An angle is ratio of two sides. Usually these two sides are of a right angle. The
relation between "the angle" and "the ratio of lengths of the two sides" is given
by use of the unit circle. The unit circle is used to calculate the sine, cosine and
other trigonometric values.



There are two methods one is a graphical method the other is an algebraic method.

We could solve this graphically by drawing to scale (e.g.1cm= 1kN) each vector
magnitude and arrange each magnitude so that Tug B vector length has Tug A
vector length starting at its end rather than at the origin of the graph.
(magnitude =length when drawing a vector)
The angle to the reference direction can be maintained by the use of a line parallel
to the reference line placed at a convenient distance along the x-axis. (in our case
the x-axis length of Tug B vector.) The resultant can be drawn and measured with
a ruler to determine the length ( a number Newtons in our problem) and protractor
to determine the angle to the line XO.

To draw in the resultant a line is drawn from the origin (point X in the diagram) to
the end of the last vector. This vector is also the longest side of a triangle as we have
two vectors being combined. this triangle has as its sides the 3 forces Tug A and
Tug B as the two shorter sides, and the resultant is the longest side.

Drawing is not easy to do on the computer so an illlustration is not given.



We could also solve this algebraically.

There are number of steps to the algebraic method
1. define x,y coordinate system. (i.e turn the diagram into a graph)
2. make sure each vector uses the same units of measurement
and keep in mind the units used.
3.choose a scale
4. convert all vectors to ax+by form.
5. sum the ax+by form of the both vectors.
6.convert the resultant vector to the form used in the statement of the problem
(N/SdegreesE/W bearing format in our problem.)



1.) The first step is to define an x,y coordinate system (i.e. draw or imagine x-axis an y-axis)

**Let the the y- axis be the line extending from Point X through Point O. (let us call it XO for
convenience). XO is drawn in the diagram in the top to bottom arrangement just like y-axis is
drawn in a x,y coordinate system.

**Let the x-axis be a line 90 degrees to XO intersect (or crossing) XO at point X.


2.) Now that we have an x,y coordinate sysem set up, we should Keep in mind that
each unit will be read by us as units of kN (kiloNewtons) as those units are used
in the diagram.

3.) the scale used in the statement of the problem is kN. (meaning multiples 1,000 N.)

4.) Now that we have a coordinate system, a scale, and we are aware of the units of
measurement we are using , we can describe the force of each boats vectors as a
sum units of x and units of y.

Using this system, the pulling force of each tug will be written in the format of
Tug A = ax+by. The letters a and b are the number of units along the axis.
So ax is the number of units along the x axis.

We need to change the vector notation from the bearings format in the diagram
to ax+by format so that we can add the portion of each force that is along the x-axis
and the portion of each force that is along the y-axis. We can consider our x,y
coordinate system to be inside a unit circle. In the unit circle the y-axis give the
sine value of an angle, and the x-axis gives the cosine of an angle.

ax is evaluated as the cosine of 'the angle from the x-axis' multiplied by the vector
magnitude in the diagrams bearing .

by is evaluted as the sine of 'the angle from the x-axis' multiplied by the vector magnitude
in the diagrams bearing.

This format change should be done for both Tug A and Tug B.

5.) Then adding the two vectors Tug A and Tug B is done as follows

Tug A.....ax+by
Tug B......cx+dy
Resultant....(a+c)x + (b+d)y let us write ex+fy.

the terms in brackets (a+c) and (b+d) can be simplified when the those terms are evaluated.

6.) The form the resultant is in is not the same as the form as the vector in the
statement of the problem so we should convert the resultant vector from the
ax+by form to N/S degreesE/W bearing form.

This has to steps A)the magnitude of the resultant vector and B) the angle.

A) Resultant magnitude = square root of (e squared added to f squared).

B) The resultant angle from x-axis = inverse tan of (f /e)
f is the vertical side, e is the horizontal side, the resultant is the hypotnuse of
a triangle inside a unit circle.
this angle is from the x-axis we need an angle from the OX line ( our y-axis)
this can be evaluate by
the resultant angle from the OX line = 90 - (inverse tan of (f /e)).


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