Vector Graphing Help: A+B & A-B

[BunDa4Th]

Help with Vector/magnitude

Vector A has a magnitude of 7.30 units and makes an angle of 46.5° counter-clockwise from the positive x-axis. Vector B has a magnitude of 8.00 units and is directed along the negative x-axis.

(a) Using graphical methods, find the vector sum A + B.
Magnitude of A + B: units
Direction of A + B: ° counterclockwise from +x-axis

(b) Using graphical methods, find the vector difference A - B.
Magnitude of A - B: units
Direction of A - B: ° counterclockwise from +x-axis

I am having trouble with that problem and I was wondering if I drew my graph correctly?

 

[radou]

http://img.search.com/e/ef/300px-Vector_addition.png"

The graphical method to measure magnitude is to a) declare 1 [cm] :: n [units] b) measure your vector to retrieve a number A c) multiply A*n to get the magnitude

 

[Astronuc]

When adding vectors, e.g. A + B, the tail of the second vector is placed at the point (head) of the first vector. The orientation with respect to the axes is preserved. Subtracting a vector is equivalent to adding a negative vector, and the negative vector is oriented 180° from the positive vector direction.

Another way to graphically draw the subtraction operation is to put the tails of the two vectors together, and the resultant vector is proceeds from the point (head) of the first vector to the head of the second vector.

 

[pavadrin]

this problem can be worked out without using a graphical representation, by breaking the vectors into i and j components using simple trigonometry

Vector A = \left( \begin{array}{l}<br /> 7.3\sin 46.5^ \circ \\ <br /> 7.3\cos 46.5^ \circ \\ <br /> \end{array} \right)

http://img171.imageshack.us/img171/5659/vectoramp1.jpg

Vector B = \left( \begin{array}{l}<br /> 8\sin 180^ \circ \\ <br /> 8\cos 180^ \circ \\ <br /> \end{array} \right)

the i and j components can now be equated:

\left( \begin{array}{l}<br /> 7.3\sin 46.5 \\ <br /> 7.3\cos 46.5 \\ <br /> \end{array} \right) + \left( \begin{array}{l}<br /> 8\sin 180 \\ <br /> 8\cos 180 \\ <br /> \end{array} \right) = \left( \begin{array}{l}<br /> 7.3\sin 46.5 + 8\sin 180 \\ <br /> 7.3\cos 46.5 + 8\cos 180 \\ <br /> \end{array} \right) = \left( \begin{array}{l}<br /> 5.2952 \\ <br /> - 2.9750 \\ <br /> \end{array} \right)

by using Pythagoras theorem a^2 = b^2 + c^2 and simple trigonometry the vector length and the direction can be worked out:

\begin{array}{c}<br /> \left| R \right| = \sqrt {5.2952^2 + \left( { - 2.975} \right)^2 } \\ <br /> = 6.0737\quad units \\ <br /> \theta = \arctan \frac{{5.2952}}{{ - 2.975}} \\ <br /> = 60.6715^ \circ \\ <br /> \end{array}

the angle is 60.6715 degrees however this value needs to be subtracted from 180 to give the value for direction as it has rotated from the x-axis

http://img105.imageshack.us/img105/733/directionad9.jpg

this method is better than a graphical representation is it will yeild more accirate results and enables more vecotrs to be added and subtracted easier. when a vector is subtracted, it is though it is being added, but in the opposite direction. the graphical representation is still helpful as it enables you to understand how vectors work.

hope this help,
Pavadrin

 

[BunDa4Th]

I am having trouble figuring this out.

One of the fastest recorded pitches in major-league baseball, thrown by
Billy Wagner in 2003, was clocked at 101.0 mi/h (Fig. P3.22). If a pitch were thrown horizontally with this velocity, how far would the ball fall vertically by the time it reached home plate, 60.5 ft away? (ft)

So far I figure I will have to use
DeltaX = V_oxT
DeltaY = V_oyT - 1/2GT^2

V_ox = 101 mi/h and DeltaX = 60.5 ft (.01mi)

If correctly I am suppose to solve for DeltaY to get the answer but everytime I do it I get a really huge number.

 

[Hootenanny]

Perhaps if you show your working we could check it; you may wish to check your conversions, all your distances, velocities and time should be in appropriate units.

 

[BunDa4Th]

68.7 = 62.7T T= 1.10

deltaY = -1/2(-9.8)(1.10)^2 = 5.93 ft

That is what I did and that is still the wrong answer.

 

[BunDa4Th]

I finally figure this out. I realize what the question was really asking.