Cheatography

# Intro to Forces Cheat Sheet (DRAFT) by pastel-galaxies

A cheat sheet explaining Newton's Three Laws of Motion, types of force, and incline mechanics.

This is a draft cheat sheet. It is a work in progress and is not finished yet.

### Newton's Three Laws

 Newton's Three Laws of Motion set up the basic parameters for why objects behave the way they do. They are vital for unders­tanding the units concerning force, work, and energy.
Newton's Three Laws are as follows:
1. INERTIA: An object at rest will remain at rest and an object in motion will remain in motion unless acted upon by an outside force.
2. DYNAMICS: The accele­ration of an object is equal to the sum of all forces acting upon it divided by its mass.
3. ACTION AND REACTION: To every action there is an equal and opposite reaction.

### Newton's Three Laws (Expla­ined)

 INERTIA Inertia is the tendency of an object to resist motion. This means that an object will not move unless it is being forced to by something (gravity, a push, a force, etc.) DYNAMICS This law basically says that F = m × a, one of the most important equations in physics. ACTION­/RE­ACTION This law explains that for every force, there is a force of the same value that opposes it. For example, if you push on a wall with 100 N of force, the wall will push back with 100 N of force.

 In a situation where something is being pulled up or down, how do I know which force is acting and which force is reacting? The force that is doing the action is the active force (will come first in the equation) and the other force will be second (will come second in the equation). Generally if something is being pulled, then tension is doing the action. How do I know when to use ΣF = 0? If the mass is in static equili­brium, which means nothing is moving. Therefore, you can set the "­up" force (tension) equal to the "­dow­n" force (gravity), and set the "­lef­t" force equal to the "­rig­ht" force. This does not apply for diagonal forces!

### Two-String Tension pt. 3

From this point forward, I've drawn in some extra triangles to make the problem easier to handle. By using geometry and some simple math, you can find the other angles in the two new triangles (I can't draw these into the image). The bottom right line is T2x, the right side is T2y, the left side is T1y, and the bottom left line is T1x.

### Tension

 Tension is entirely a type of force, but it's best to treat it as a separate idea from things like FG or FN. In a situation where an object is hanging from the ceiling, tension works as an opposition to the force of gravity.

### Tension Explained

In this situation, to solve for tension, we would use the equation F = m × a, but since we are looking for net force, we would have to use FG - T, because both forces are acting on the box.

### Tension and SOHCAHTOA

 Unders­tanding SOHCAHTOA and how to use it is incredibly important in terms of tension and force (espec­ially on inclines). SOHCAHTOA is a way to find angles and sides of a triangle using parts of it that you know. In any triangle, knowing how to find missing angles and sides is crucial to solving problems.
SOHCAHTOA stands for sine = opposi­te/­hyp­ote­nuse, cosine = adjace­nt/­hyp­ote­nuse, tangent = opposi­te/­adj­acent. Sine, cosine and tangent are often shortened to sin, cos and tan.

### Two-String Tension

Find T1 and T2 if w = 435 kg.

This problem has a lot of steps, so it's easier to explain them as I go along.
First, we can find FG, which is mass × gravity. Since the problem tells us that mass is 435 kg, we know that FG = 4263 N.

Continue on to the second box.

### Example Problem #1

 How much force is needed to accelerate a 70 kg skier at 3 m/s2? This question is pretty straig­htf­orward - all you need to do is use the equation F = ma and plug in 70 for the mass and 3 for the accele­ration.
The answer to this question should be 210 N.

### Example Problem #2

 If forces of 30 N and 35 N act in opposite directions on a 40 kg object, what is the accele­ration of the object? This question is pretty similar to the first, except that now we need to find the net force - the sum of all the forces acting on the object. Be careful - usually sum means addition, but because the forces go in opposite direct­ions, we need to subtract them. Therefore, 35-30 = 5, then we plug in 5 N for the force and 40 for the mass.
The answer to this problem should be 0.125 m/s2.

### Example Problem #3

 Imagine a light hanging from the ceiling by a rope. The mass of the light is 50 kilograms. A) what are the two forces involved in this situation, B) what is the accele­ration of the light, and C) find the tension in the string. Believe it or not, this is all we need to solve this problem. Again, we are going to use the equation F = m × a, but we need to change it to FG - T = m × a because we need the sum of the forces acting upon the light. Since we know FG, we know m, and we know a, we can plug them in and solve for T.
The answers to this problem should be: A) tension and force due to gravity, B) accele­ration = 0, and C) T = 490 N.

### Two-String Tension pt. 2

 Next, we can say that ΣFy = 0 (because nothing on the y-axis is moving). Since there's three forces on the y-axis - T1y, T2y, and FG - we can break this formula down into (T1y + T2y) - FG = 0. By doing some simple algebra, we can rewrite this as T1y + T2y = 4263 N. We'll set this formula aside and look at the x-axis. We know that ΣFx = 0 because there's no movement on the x-axis either. Since there's only 2 forces on the x-axis, we can say that T2x - T1x = 0, and we can rewrite that as T2x = T1x. Continue to the third box.

### Two-String Tension pt. 4

 This is where SOHCAHTOA comes in. First, we know that cos 53 = T2x/T2. By multip­lying both sides by T2, we get T2 cos 53 = T2x. By the same principle, cos 37 = T1x/T1 and T1 cos 37 = T1x. Then, since T2x = T1x, set T2 cos 53 = T1 cos 37. Divide both sides by cos 37 to get T1 = 0.75T2. Now, we're going to go back to the y formulas. Using SOHCAHTOA, T1y = T1 sin 37 and T2y = T2 sin 53. Since we have substitute values for T1y and T2y, we can plug them into our original y equation: (T1y + T2y) = 4263. By using our new values, we get that T1 sin 37 + T2 sin 53 = 4263. Then, we substitute THOSE values for T1 = 0.75T2 and we get the equation (0.75T2) sin 37 + T2 sin 53 = 4263, which simplifies to 1.25T2 = 4263. Our last step is to divide both sides by 1.25 and you have a value for T2. Then, just put that value into the equation T1 = 0.75T2.
The answers to this problem should be T1 = 2557.8 N and T2 = 3410.4 N.