03.1

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Problem One

Find the components of this vector:

PQ3.1 1.gif A=_________x + _________y


To solve this problem you use:
A_x\ = (mag)cos\Theta
A_y\ = (mag)sin\Theta


To solve:
A_x\ = (32)cos17
A_x\ = 30.60 m\hat{x}


A_y\ = (32)sin17
A_y\ = 9.35 m\hat{y}

A\ = 30.60 m\hat{x}\ + 9.356 m\hat{y}

Problem Two

Find the components of this vector:

PQ3.1 2.2.GIF B=_________x + _________y

Again:
A_x\ = (mag)cos\Theta
A_y\ = (mag)sin\Theta


So, we do the same thing as before. This time, don't forget that \Theta\, must always be a trig angle!
 \; 270^\circ\; + 23^\circ = 293^\circ

A_x\ = (15)cos293
A_x\ = 5.861 m\hat{x}


A_y\ = (15)sin293
A_y\ = -13.81 m\hat{y}

B\ = 5.861 m\hat{x}\ + -13.81 m\hat{y}

Problem Three

Convert this vector component vector to an angle magnitude vector. Draw it as an arrow, and label any angle it makes with its value, and find the magnitude.

C = 13.2 m/s x + 5.70 m/s y

We solve this problem like the first two, only in reverse. The first thing that you do is draw out the equation without lifting your pen. It should look like this:

Pq3.1.GIF



Using SOHCAHTOA, we can take the inverse tangent (the arctan) of these values to find the angle, and using the pythagorean theorem we can take the square root of x squared plus y squared to get the magnitude.

 Mag = \sqrt{(x^2 + y^2)}
 Mag = \sqrt{(5.70^2 + 13.2^2)}

Mag = 14.4 m/s

Angle = \arctan\,{(\frac{O}{A})}

Angle = \arctan\,{(\frac{5.70}{13.2})}
Angle = 23.4o

Angle Magnitude: 14.4 m/s @ 23.4o

Problem Four

Given:

D\ = 16.2 m\hat{x}\ + -3.5 m\hat{y}

A\ = -13.7 m\hat{x}\ + -4.2 m\hat{y}


Find:
D+E=_________x + _________y
E-D=_________x + _________y

To solve this problem, you add and subtract the vector components like basic arithmetic.

Finding D+E

With this problem, you simply add the corresponding values.
 +16.2 m\hat{x}\ + -3.5 m\hat{y}

 -13.7 m\hat{x}\ + -4.2 m\hat{y}
+___________________________
 +2.5 m\hat{x}\ + -7.7 m\hat{y}


D+E\ = 2.5 m\hat{x}\ + -7.7 m\hat{y}

Finding E-D

As with the previous problem, you subtract the D values from the E values to find the answer.
 -13.7 m\hat{x}\ + -4.2 m\hat{y}
 +16.2 m\hat{x}\ + -3.5 m\hat{y}
-___________________________
 -29.9 m\hat{x}\ + -.7 m\hat{y}


E-D\ = -29.9 m\hat{x}\ + -.7 m\hat{y}


Problem Five

Calculate the sum of vectors A and B (From problem 1 and 2). Draw the resultant vector as an angle magnitude vector. (Label your angle and magnitude clearly – use an arrow) Do your work on the back of this sheet, and label and describe the three steps for doing so.

Sum of vectors

Calculate the sum of the two vectors

  30.6 m\hat{x}\ +   9.356 m\hat{y}

 5.861 m\hat{x}\ + -13.81 m\hat{y}
+___________________________
 36.461 m\hat{x}\ + -4.454 m\hat{y}


Draw the vector

Label the angle, sides, and vector.

Pq3.15.GIF

Calculate the Angle Magnitude

Do the inverse tangent and square root of the two distances squared. Write these values as angle magnitude.

 Mag = \sqrt{(x^2 + y^2)}
 Mag = \sqrt{(4.46^2 + 36.461^2)}

Mag = 36.8 m/s

Angle = \arctan\,{(\frac{O}{A})}

Angle = \arctan\,{(\frac{4.46}{36.461})}
Angle = 6.97o

Angle Magnitude: 36.8 m/s @ 6.97o