Chapter 3 Part 2 Right Triangles, Analytic Addition of Vectors, Projectile Motion

Prof. Clements Notes/Hints for Physics Courses

Chapter 3 Open Stax College Physics or Most Introductory Physics Courses

Right Triangles, Analytic Addition of Vectors, Projectile Motion

I highly recommend that you go to the web site https://openstax.org/details/books/college-physics to access the free OpenStax College Physics textbook and resources.

YouTube videos of my (usually short) lectures and example problems are indexed at http://www.physics.gpclements.com/ . There are also a few videos that give review of basic math tools that are used in an introductory physics course.

Terms you should know at the end of this unit: analytic addition, component, hypotenuse, projectile motion

Vector Addition and Subtraction Using Analytical Methods

Analytic addition of vectors produces accurate results. This is the preferred method for adding vectors. Having said that, it is worth your time to still make a quick graphical solution to all vector addition problems before you apply analytical methods. This will help you catch errors in your calculations.

In the typical problem you have two vectors that are specified with their magnitude (length) and angle to the positive X axis given. In the analytical method you will break the vectors into components using sine and cosine, add the X components, add the Y components, and then find the resultant vector's magnitude and angle.

Be careful in your thinking if you see Ax + Ay = A in a textbook. The bold font should remind you that these quantities are vectors and must be added using the rules for adding vectors, not the rules of algebra you are used to using with numbers. It is never allowed to do a simple addition of the X and Y component values so it might be better to call Ax + Ay = A vector reconstruction rather than vector addition. You will use the X and Y components to find the resultant vector, A. The magnitude of A is found using the Pythagorean theorem and the angle of the vector is found by using the inverse tangent function and your brain. (i.e. The inverse tangent function does not always give the right result.

Analytical (mathematical) addition of vectors ( C = A + B ) is performed by

1) finding the components of each vector A and B,

2) adding “like” components i.e. C_X = A_X + B_X , C_y = A_y + B_y

3) reconstructing C.

The image below gives a review of Right Triangles, sine, cosine, tangent, and the Pythagorean Theorem.

To perform the mathematical addition of vectors we must be able to resolve vectors into their components. This is accomplished by the trigonometric calculations for a right triangle. The vector is always the hypotenuse in the right triangle. The X and Y components are the sides of the right triangle. If the angle is given in standard form (i.e. measured counterclockwise from the positive X axis) then you can always use the relationships below to find the components of the vector. If A is the magnitude of the vector then

A_X = A cos (theta) and A_Y = A sin (theta) .

Suppose a certain vector has a magnitude of 12 meters and a direction of 37 degrees from the positive X axis. What are the values of the components of the vector?

A_X = 12 cos(37^o) → 9.58 meters A_y = 12 sin(37^o) → 7.22 meters

If you are given the components of a vector you can reconstruct the vector using the Pythagorean Theorem and the tangent function (plus thinking).

( ) ^½ represents square root

A = ( A_X² + A_Y² ) ^½      tan(theta) = A_Y /A_X    (theta) is found by applying tan^-1 to both sides of this relationship. The angle found will only be correct if the vector is in the first or fourth quadrant (X component is positive). If the X component is negative you must adjust the angle by 180^o . Your sketch of the graphical addition of the two vectors should help you find the correct angle.

Perform the reconstruction of A using the A_X and A_Y values given above.

A = ( (9.58)² + (7.22)² )^½ → 11.996 or 12 meters

tan(theta) = A_Y /A_X      tan^-1 ( tan(theta) ) = tan^-1 (A_Y /A_X) →  (theta) = tan^-1 (7.22 / 9.58) →  (theta) = 37^o

If your calculator does not produce this result for the angle check that the mode of the calculator is set for degrees, not radians.

Suppose A has a length of 3 meters and an angle of 30 degrees, B has a length of 7 meters and an

angle of 160 degrees. Find C = A + B.

A_X = 3 cos(30^o) → 2.598 meters A_y = 3 sin(30^o) → 1.5 meters

B_X = 7 cos(160^o) → -6.578 meters B_y = 7 sin(160^o) → 2.394 meters

C_X = -3.9798 C_y = 3.894

C = ( (-3.9798)² + (3.894)² )^½ → 5.57 meters

tan(theta) = A_Y /A_X     (theta) = tan^-1 (3.894 / -3.9798) → (theta) = -44.4^o but this angle is incorrect.  The result vector, C, is in the second quadrant.

In this situation you must add 180^o to -44.4^o to produce the correct value 

for (theta) of 135.6^o .

Vector addition can be extended to any number of vectors. The following image shows how to add three vectors.

When a problem has two dimensions you must work the problem using vector techniques. Make a vector diagram before you attempt to solve the problem.

You should work some example problems and let your instructor know if you have questions.

Projectile Motion

Displacement, velocity, and acceleration are vectors. When motion occurs in a direction not aligned with either the X or Y axis you must work two problems simultaneously. It is very important that you keep all X quantities (horizontal motion) in one equation and all Y quantities (vertical motion) in a separate equation. The only quantity which will be common to both equations is the time value, t.

Your result will have X and Y components of the position, velocity, and acceleration. You will need to reconstruct the vector to determine the magnitude and angle of the displacement or velocity or acceleration.

Projectile motion refers to motion which takes place near the surface of a large mass (e.g. the earth).

Carefully toss a soft object to another person and have the person throw the object back to you. Watch the object carefully as it flies through the air. The object is undergoing projectile motion. In projectile motion the vertical acceleration is -9.80 m/s² (assuming the object is moving near the earth, not Mars etc.) and the value of the horizontal acceleration is 0! (As long as we ignore air resistance.)

Recall that acceleration determines how the velocity changes. Because the horizontal acceleration is 0 the horizontal velocity is constant!

Solving Projectile Motion Problems

Because the acceleration in the X (horizontal) direction is zero the kinematic equations for the X motion simplify to distance = (initial X component of velocity) * time. The Y (vertical) equations will contain -9.8 m/s² for the acceleration value.

It is very important that you sketch the problem in projectile motion problems. You must choose a positive direction for the horizontal and vertical motion. I will choose upward to be the positive direction. You must choose the location for 0 for both the horizontal and vertical coordinates. I will usually choose ground level to be the 0 location for the X and Y values.

Here is the link for the PhET Cannon projectile motion simulation. Click the “Play” button arrow on the graphic to run the simulation. Your computer will need to have JAVA installed.

PhET Cannon

http://phet.colorado.edu/en/simulation/projectile-motion

The following projectile motion example shows how to calculate the 1) maximum height, 2) time in the air, 3) landing location.

The following projectile motion example analyzes the motion of a basketball.

The following projectile motion example analyzes a situation in which two balls collide.

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