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5.21: Resultant as Magnitude and Direction

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You are out playing soccer with friends after school one day when you and a friend kick the soccer ball at the same time. However, you kick the ball with 55 N of force with a vector like this:

and your friend kicks the ball with 70 N of force with a vector like this:

The angle between the two vectors is 74^\circ , and combined the graph of both vectors looks like this:

Can you represent the net force on the ball? What will the resultant vector be like?

Keep reading, and by the time you finish this Concept, you'll be able to make this calculation and draw the resultant vector.

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Applications of Vector Addition

Guidance

If we don’t have two perpendicular vectors, we can still find the magnitude and direction of the resultant without a graphic estimate with a construction using a compass and ruler. This can be accomplished using both the Law of Sines and the Law of Cosines.

Let's investigate some examples of this.

Example A

\vec{A} makes a 54^\circ angle with \vec{B} . The magnitude of \vec{A} is 13.2. The magnitude of \vec{B} is 16.7. Find the magnitude and direction the resultant makes with the smaller vector.

There is no preferred orientation such as a compass direction or any necessary use of x and y coordinates. The problem can be solved without the use of unit vectors.

Solution: In order to solve this problem, we will need to use the parallelogram method. Since vectors only have magnitude and direction, one can move them on the plane to any position one wishes, as long as the magnitude and direction remain the same. First, we will complete the parallelogram: Label the vectors. Move \vec{b} so its tail is on the tip of \vec{a} . Move \vec{a} so its tail is on the tip of \vec{b} . This makes a parallelogram because the angles did not change during the translation. Put in labels for the vertices of the parallelogram.

Since opposite angles in a parallelogram are congruent, we can find \angle A .

\angle{CBD} +  \angle{CAD} + \angle{ACB} + \angle{BDA} & = 360 \\2 \angle{CBD} + 2 \angle{ACB} & = 360 \\\angle{ACB} & = 54^\circ \\2 \angle{CBD} & = 360 - 2(54) \\\angle{CBD} & = \frac{360-2(54)}{2} = 126

Now, we know two sides and the included angle in an oblique triangle. This means we can use the Law of Cosines to find the magnitude of our resultant.

x^2 & = 13.2^2 + 16.7^2 - 2(13.2)(16.7) \cos 126 \\x^2 & = 712.272762  \\x & = 26.7

To find the direction, we can use the Law of Sines since we now know an angle and a side across from it. We choose the Law of Sines because it is a proportion and less computationally intense than the Law of Cosines.

\frac{\sin \theta}{16.7} & = \frac{\sin 126}{26.7} \\\sin \theta & = \frac{16.7 \sin 126}{26.7} \\\sin \theta & = 0.5060143748 \\\theta & = \sin ^{-1} \ 0.5060 = 30.4^\circ

The magnitude of the resultant is 26.7 and the direction it makes with the smaller vector is 30.4^\circ counterclockwise.

We can use a similar method to add three or more vectors.

Example B

Vector A makes a 45^\circ angle with the horizontal and has a magnitude of 3. Vector B makes a 25^\circ angle with the horizontal and has a magnitude of 5. Vector C makes a 65^\circ angle with the horizontal and has a magnitude of 2. Find the magnitude and direction (with the horizontal) of the resultant of all three vectors.

Solution: To begin this problem, we will find the resultant using Vector A and Vector B . We will do this using the parallelogram method like we did above.

Since Vector A makes a 45^\circ angle with the horizontal and Vector B makes a 25^\circ angle with the horizontal, we know that the angle between the two (\angle{ADB}) is 20^\circ .

To find \angle{DBE} :

2 \angle{ADB} + 2 \angle{DBE} & = 360 \\\angle{ADB} & = 20^\circ \\2 \angle{DBE} & = 360 - 2(20) \\ \angle{DBE} & = \frac{360-2(20)}{2} = 160

Now, we will use the Law of Cosines to find the magnitude of DE .

DE^2 & = 3^2 + 5^2 - 2(3)(5) \cos 160 \\DE^2 & = 62 \\ DE & = 7.9

Next, we will use the Law of Sines to find the measure of \angle EDB .

\frac{\sin 160}{7.9} & = \frac{\sin \angle{EDB}}{3} \\\sin \angle{EDB} & = \frac{3 \sin 160}{7.9} \\\sin \angle{EDB} & = .1299 \\\angle{EDB} & = \sin^{-1} \ 0.1299 = 7.46^\circ

We know that Vector B forms a 25^\circ angle with the horizontal so we add that value to the measure of \angle{EDB} to find the angle DE makes with the horizontal. Therefore, DE makes a 32.46^\circ angle with the horizontal.

Next, we will take DE , and we will find the resultant vector of DE and Vector C from above. We will repeat the same process we used above.

Vector C makes a 65^\circ angle with the horizontal and DE makes a 32^\circ angle with the horizontal. This means that the angle between the two (\angle{CDE}) is 33^\circ . We will use this information to find the measure of \angle{DEF} .

2\angle{CDE} + 2\angle{DEF} & = 360 \\ \angle{CDE} & = 33^\circ \\2 \angle{DEF} & = 360 - 2(33) \\\angle{DEF} & = \frac{360-2(33)}{2} = 147

Now we will use the Law of Cosines to find the magnitude of DF .

DF^2 & = 7.9^2 + 2^2 - 2(7.9)(2) \cos 147 \\DF^2 & = 92.9 \\DF & = 9.6

Next, we will use the Law of Sines to find \angle{FDE} .

\frac{\sin 147}{9.6} & = \frac{\sin \angle{FDE}}{2} \\\sin \angle{FDE} & = \frac{2 \sin 147}{9.6} \\\sin \angle{FDE} & = .1135 \\\angle{FDE} & = \sin^{-1} 0.1135 = 6.5^\circ = 7^\circ

Finally, we will take the measure of \angle{FDE} and add it to the 32^\circ angle that DE forms with the horizontal. Therefore, DF forms a 39^\circ angle with the horizontal.

Example C

Two forces of 310 lbs and 460 lbs are acting on an object. The angle between the two forces is 61.3^\circ . What is the magnitude of the resultant? What angle does the resultant make with the smaller force?

Solution: We do not need unit vectors here as there is no preferred direction like a compass direction or a specific axis. First, to find the magnitude we will need to figure out the other angle in our parallelogram.

2 \angle{ACB} + 2 \angle{CAD} & = 360 \\\angle{ACB} & = 61.3^\circ \\2 \angle{CAD} & = 360 - 2(61.3) \\\angle{CAD} & = \frac{360 - 2 (61.3)}{2} = 118.7

Now that we know the other angle, we can find the magnitude using the Law of Cosines.

x^2 & = 460^2 + 310^2 - 2(460)(310) \cos 118.7^\circ \\x^2 & = 444659.7415  \\x & = 667

To find the angle the resultant makes with the smaller force, we will use the Law of Sines.

\frac{\sin \theta}{460} & = \frac{\sin 118.7}{666.8} \\\sin \theta & = \frac{460 \sin 118.7}{666.8} \\\sin \theta & = .6049283888 \\\theta & = \sin^{-1}\ 0.6049 = 37.2^\circ

Vocabulary

Resultant: The resultant is a vector that is the sum of two or more vectors.

Guided Practice

1. Forces of 140 Newtons and 186 Newtons act on an object. The angle between the forces is 43^\circ . Find the magnitude of the resultant and the angle it makes with the larger force.

2. An airplane is traveling at a speed of 155 km/h. It's heading is set at 83^\circ while there is a 42.0 km/h wind from 305^\circ . What is the airplane's actual heading?

3. If \overrightarrow{AB} is any vector, what is \overrightarrow{AB} + \overrightarrow{BA} ?

Solutions:

1. magnitude = 304, 18.3^\circ between resultant and larger force

2. Recall that headings and angles in triangles are complementary. So, an 83^\circ heading translates to 7^\circ from the horizontal. Adding that to 35^\circ ( 270^\circ from 305^\circ ) we get 42^\circ for two of the angles in the parallelogram. So, the other angles in the parallelogram measure 138^\circ each, \frac{360-2(42)}{2} . Using 138^\circ in the Law of Cosines, we can find the diagonal or resultant, x^2 = 42^2 + 155^2 - 2(42)(155) \cos 138 , so x = 188.3 . We then need to find the angle between the resultant and the speed using the Law of Sines. \frac{\sin a}{42} = \frac{\sin 138}{188.3} , so a = 8.6^\circ . To find the actual heading, this number needs to be added to 83^\circ , getting 91.6^\circ .

3. BA is the same vector as AB , but because it starts with B it is in the opposite direction. Therefore, when you add the two together, you will get (0,0).

Concept Problem Solution

As you've seen in this Concept, you can represent the vector resulting from both of your forces as the resultant of vector addition. Since vectors only have magnitude and direction, one can move them on the plane to any position one wishes, as long as the magnitude and direction remain the same. First, we will complete the parallelogram: Label the vectors. Move \vec{b} so its tail is on the tip of \vec{a} . Move \vec{a} so its tail is on the tip of \vec{b} . This makes a parallelogram because the angles did not change during the translation. Put in labels for the vertices of the parallelogram.

Since opposite angles in a parallelogram are congruent, we can find angle A .

\angle{CBD} +  \angle{CAD} + \angle{ACB} + \angle{BDA} & = 360 \\2 \angle{CBD} + 2 \angle{ACB} & = 360 \\\angle{ACB} & = 72^\circ \\2 \angle{CBD} & = 360 - 2(72) \\\angle{CBD} & = \frac{360-2(72)}{2} = 108

Now, we know two sides and the included angle in an oblique triangle. This means we can use the Law of Cosines to find the magnitude of our resultant.

x^2 & = 70^2 + 55^2 - 2(70)(55) \cos 72 \\x^2 & = 5545.569  \\x & = 74.47

To find the direction, we can use the Law of Sines since we now know an angle and a side across from it. We choose the Law of Sines because it is a proportion and less computationally intense than the Law of Cosines.

\frac{\sin \theta}{70} & = \frac{\sin 108}{74.47} \\\sin \theta & = \frac{70 \sin 108}{74.47} \\\sin \theta & = 0.89397 \\\theta & = \sin ^{-1} \ 0.89397 = 63.68^\circ

The combined force of your kick along with your friend's is 74.47 Newtons and the direction it makes with your kick is 63.68^\circ counterclockwise.

Practice

\vec{a} makes a 42^\circ angle with \vec{b} . The magnitude of \vec{a} is 15. The magnitude of \vec{b} is 22.

  1. Find the magnitude of the resultant.
  2. Find the angle of the resultant makes with the smaller vector.

\vec{c} makes a 80^\circ angle with \vec{d} . The magnitude of \vec{c} is 70. The magnitude of \vec{d} is 45.

  1. Find the magnitude of the resultant.
  2. Find the angle of the resultant makes with the smaller vector.

\vec{e} makes a 50^\circ angle with \vec{f} . The magnitude of \vec{e} is 32. The magnitude of \vec{f} is 10.

  1. Find the magnitude of the resultant.
  2. Find the angle of the resultant makes with the smaller vector.

\vec{g} makes a 100^\circ angle with \vec{h} . The magnitude of \vec{g} is 50. The magnitude of \vec{h} is 35.

  1. Find the magnitude of the resultant.
  2. Find the angle of the resultant makes with the smaller vector.
  3. Two forces of 100 lbs and 120 lbs are acting on an object. The angle between the two forces is 50^\circ . What is the magnitude of the resultant?
  4. Using the information from the previous problem, what angle does the resultant make with the larger force?
  5. A force of 50 lbs acts on an object at an angle of 30^\circ . A second force of 75 lbs acts on the object at an angle of -10^\circ . What is the magnitude of the resultant force?
  6. Using the information from the previous problem, what is the direction of the resultant force?
  7. A plane is flying on a bearing of 30^\circ at a speed of 450 mph. A wind is blowing with the bearing 200^\circ at 50 mph. What is the plane's actual direction?
  8. Vector A makes a 30^\circ angle with the horizontal and has a magnitude of 4. Vector B makes a 55^\circ angle with the horizontal and has a magnitude of 6. Vector C makes a 75^\circ angle with the horizontal and has a magnitude of 3. Find the magnitude and direction (with the horizontal) of the resultant of all three vectors.
  9. Vector A makes a 12^\circ angle with the horizontal and has a magnitude of 5. Vector B makes a 25^\circ angle with the horizontal and has a magnitude of 2. Vector C makes a 60^\circ angle with the horizontal and has a magnitude of 7. Find the magnitude and direction (with the horizontal) of the resultant of all three vectors.

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Date Created:

Sep 26, 2012

Last Modified:

May 27, 2014
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