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Trigonometry and Coordinate systems

After you have a free body diagram, the next step to every physics problem is to turn that information into an equation based on either Newton's 2nd law, F=ma, or plug it into the energy equation

F=ma applies independently in each direction, with F being the sum of forces in one of the directions. Summing the components in both x and y axes is generally a good idea, but often the direction of motion (DOM) is the only one needed. For components that move in only one direction, like in Atwood's machine, 1D kinematics, and Block on a slope, I "assign" whatever direction I think it will move as positive (gutcheck only, if wrong you'll get a negative answer which is ok). Some problems have motion in both x and y at the same time but in a straight line, such as the top-down box problem, and assigning the direction of the net force as positive DOM makes the later steps significantly easier. For any problem with friction, you will probably need to find the sum of forces in the direction perpendicular to the DOM, the normal direction (preferably 90 deg CCW), before you can solve the DOM equation. without friction, the normal direction is sometimes not required, so do it second after checking if the DOM F=ma can solve the equation. If the normal force is entirely pointing in the normal direction (or x/y axes) the sum of forces is 0 in that direction, with the important caveat that this refers to the relative acceleration between 2 contacting objects, but not an outside observer, see the elevator problem)

To find the components of each force in each direction we need a little bit of trig. SOH-CAH-TOA is a mnemonic device that is used in math to remember the definitions of the three most common trigonometric functions, for engineers and physics majors, it's worth memorizing for later classes. Using them for every problem individually will work, but it's slow, especially if the reference angle (the dark corner in the image below) for each force is off of a different axis; later, I'll show my process to do it faster and foolproof.


The magnitude of the force lies on the hypotenuse of the triangle, and the components are the bottom and far side (adjacent and opposite). F*sin(theta) = the magnitude in the direction parallel to the opposite side. depending on what the angle is referenced to, this could be the x or y-axis, and it could be a negative or positive force, if it's the same direction as the positive axis, it's a positive force. The adjacent side can be found with F*cos(theta), but still needs a sign check. (example further down).


This method is useful for 3-4-5 triangles (angle 37 deg), or angles on the unit circle, where using a calculator can be avoided

If this sounds annoying, it is, and is the largest source of mistakes I see my students make, either swapping sin and cos, or forgetting to do the sign check.

A common problem is demonstrated below in how to convert a FBD into F=ma.


Luckily there is an easier way. First, find the angle the vector makes with the positive x-axis, with CCW positive in a right hanged cordinate system. It is acceptable to measure a 330 deg angle as -30 deg. Now finding the components, and direction is as simple as,

F_x = F*cos(theta)

F_y = F*sin(theta)

Your calculator will automatically output a negative value if the force points left or down, as long as the angle is input correctly. It's worth testing what inputs will result in the same value to gain some intuition for these problems. For example,

cos(120) = -sin(30), or cos(37) = cos(-37), or sin(10) = -sin(-10)

note that F is always assumed positive, and if it's negative it's the same as adding 180 deg to the angle.

This works perfectly fine in radians but made sure your calculator is in the right setting

(for the TI series, MODE - 4th row, Desmos defaults to radians)

After some practice with these, look up some trig identities, and try proving them geometrically 

The same problem as above is done agian with the easiaer method. The lift vector uses the easy way, by using the absolute ange (CCW from +x to +y) and allows the trig functions to output the sign (this can be done on the unit circle for these angles). The weight acts only in the -y dircetion, so its only included on the F_y equation, and has a negitive sign in front of it.


Two more tools needed for the basics of trig are the Pythagorean theorem and tangent. where,

F = sqrt(F_x^2 + F_y^2)

note that this syntax is (probably) accepted by mastering physics and WebAssign

where the magnitude of the force can be found given both components or can be rearranged to find the magnitude of one component (but not sign direction) given total magnitude  and the other component

F_y = sqrt(F^2 - F_x^2)

F_x = sqrt(F^2 - F_y^2)

Once you have found the magnitude and components of a vector, many questions want a final answer regarding the direction of the vector. using TOA we can find that, tan(theta) = op/aj => F_y/F_x  quite literally "rise over run" or the slope of the line.

the angle theta is then

theta = arctan(F_y/F_x)

most calculators have a tan^-1 button, this means arctan or atan, but is NOT 1/tan

But note that your calculator will output a value between 90 and -90 deg, as it cannot tell the difference between (-1)/(2) and (1)/(-2) which are angles in the 4th and 2nd quadrants respectfully or (3)/(5) and (-3)/(-5) as it cancels the negative sign. However, all of these errors are 180 deg apart, manually checking whether the final angle is in the right quadrant is an important step.

Many programming languages such as MATLAB and Excel have functions called atan2, which accept 2 inputs as the x and y coordinates (not y then x) and does the check for you.

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