free body diagrams

physics doesn’t have to suck: how to enjoy and do well in your required physics classes

As someone who doesn’t intend to take a physics class ever again, I was relieved when I walked out of my second semester physics final. That said, physics doesn’t have to suck or drag your average down. 

(1) How to enjoy physics: Adjust your attitude. Physics is so cool if you actually think about it. Your attitude will dictate your experience. (2) But physics is so hard: Change the way you study and don’t give up. I did better in university physics than in high school. The content was way more difficult but it was my studying methods that made the difference.

This post is split into 3 parts: Introductory physics (very basic physics, that unit of physics you had to do in a lower level science class), high school physics (physics from an algebra-based perspective), and university physics (calculus-based physics and labs). (Obviously these overlap a lot but I needed to organize this somehow)


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B.A.P as things my physics class said today:

Yongguk- “Don’t take shortcuts, work hard, figure out a plan. Otherwise you’ll be completely lost.”

Himchan- “I’m not telling you, you have to tell me. It’s the only way you’ll get better.” 

Daehyun- “Sorry I didn’t mean to yell I just got excited over free body diagrams.” (More like Youngjae’s free body diagram ayyyy)

Youngjae- “I’m a queen, I don’t conform to the forces of gravity. I’ll be newtons second law, you need me to solve the problem and am useless without me.”

Jongup- “I’m definitely guilty of doing too much in my head and never saying anything.”

Zelo- “I feel like this is fairly simple, I just have no idea what to do.”                         

anonymous asked:

I am seriously struggling in general physics. Non calculus based. I can't seem to grasp when to use certain formulas, and I go to office hours and tutoring and I feel I still don't know enough. What are some of your tips and tricks to retain information and learning how/when to use formulas

hi there!

so i’m quite terrible at anything related to math, including physics, so i definitely struggled through the classes, but i found a method of working with physics/formulas in general so lets see if it can work for you.

when given a word problem or scenario, i write down all the given variables, and then i write down the unknown variable that i need to solve, and from there it’s a matter of matching up what formula will work. 

for example: 

An object with a mass of 17.15 kg experiences a force of 10.99 N. What is the acceleration of the object?

Known variables:
mass = 17.15 kg
force = 10.99 N

Unknown variable to solve:
acceleration = ?

So now i go back to my list of formulas and look for ones that have mass, force, and acceleration. And f=ma pops out, and some algebra changes it to a=f/m so

a = 10.99 N/17.15 kg = 0.64 m/s/s

The tricky part about this process is if the question gives you more variables than necessary, or if you need more than one formula (for example, if figuring out acceleration was the first part of another larger problem set). 

in general the best way to learn solving problem sets is to just keep practicing until you recognize patterns in the questions and a method of solving the problems. don’t worry about memorizing formulas, because eventually after using them over and over you’ll naturally start to remember them. see if you can find practice problems online or if your professor has some. 

it also helps to learn the reason/history behind the formulas, so it’s much more than just letters and symbols. i’ve found that a full understanding of formulas (or any concept) really helps guide my problem solving. it also helps in understanding the units, like why the units N/kg would result in m/s/s in the above example.

drawing free body diagrams also helps, especially if you’re a visual learner (like me!) it also brings the “real world application” part out (like if i pretended the above question was regarding a plane or something). 

the internet also has a lot of helpful resources that gives you the option of having something taught in different styles (until one style clicks). ex. i found this website after a quick google search: Zona Land Education. youtube videos are also great. 

those are things that have worked for me. if anyone else has anything more to add, please do!! (and tagging @roundlittledog bc he’s a badass and super helpful physicist)

good luck anon!

I don’t know guys

*Opens drawer*

The MBTI stereotypes are funny, sometimes

*Pulls out erlenmeyer flask*

but honestly i just don’t see how some of them come to be

*pulls goggles out of pocket*

i mean i am an intp, right?

*conducts an electrophoresis experiment on the side*

but that doesn’t mean i like science!

*dumps chemicals from test tubes into flask*

i like a lot of other things, too!

*mixes it up with a glass rod*

just gonna let that sit for a second.

*walks over to chalkboard scrawled with hardly legible theories*

lemme just get this outta here.

*erases it all, but leaves chalk dust everywhere*

i really like to draw, actually! check this out

*draws a free-body-diagram*

you see, this arrow represents the force of gravity on the object while–

*flask suddenly explodes*

ah drat. it appears i have accidentally created another bomb.


Kinematics help

Cont’d from “Kinematics example”.

This post is best viewed on my blog, click here to be taken to its permalink.

Different problems involving kinematics often share common properites which can be used (or not used) depending on the scenario. Once you become more adept at solving these problems individually, you’ll begin to easily recognise situations where some conditions apply and others may not and start to adapt the conditions to fit the scenario. A bit of thought is required to decide logically whether one condition may be physicaly viable in the situation or not.

In this post, I will look at topics that are not strictly kinematics but have applications within the field and in wider mechanics.

Boundary conditions

There are some common boundary conditions (i.e. what happens at the extreme-most points of the motion) which come from logical reasoning.

For motion of an object with its weight perpendicular to flight motion it can be assumed that the x-component of velocity stays constant throughout the flight. This means that the final xvelocity, vx, is equal to the initial x velocity, v0, x.

The maximum height of flight, ymax,is always the point where the y velocity, vy = 0. This is a direct result of calculus, wherein dy/dt = 0 at a turning point (a maximum or minimum of a function). Now, if the time to reach the maximum height is equal to the time to reach the ground again (i.e. for perfect parabolic flight where the launch height = landing height) then we can assume that the time to reach ymax is equal to half of the full flight time.

Additionally, if we set the original position of the object at the origin of our system of co-ordinates, the point at which we find it at the same height again (the point where it lands, assuming it lands at the same height) by setting y = 0.

From there we can select an appropriate SUVAT equation.

Forces and vector resolution

See “Resolving vectors.

Sometimes you may have to combine energy conservation, momentum or force superposition (mostly, for equibrium) into your answer.

Now, if something is accelerating then its vectorial sum of all forces is equal to ma, where a is its resultant acceleration. If it is in equilibrium (“stationary” or “moving at constant velocity”) then the vector sum of its forces is equal to zero. This is a very important consequence and can be used extensively to solve problems.

You should always remember to resolve all forces/vectors in a free body diagram when working on the x and y components separately. Unless a vector is at 90° from the direction you’re resolving it to it will contribute that component. For example, the vector W acts only in the y-direction, meaning we can leave it out of our x-component resolution.

Remember, if a vector points in the opposite direction to the direction you’re working in, give its magnitude a negative sign. This is what’s intended by the vectorial (or algebraic) sum as opposed to a simply scalar summation.

Resistive forces and their applications

See a question about friction.

Generally, in kinematics we ignore the effects of resistive forces – especially air resistance – since they can make problems more complicated to solve. However, resistive forces such as the friction force Ff or the reaction force R can be used to our advantage.

The magnitude of the frictional is given by

Ff = μR,

where μ is the friction coefficient. Vectorially, we define Ff to be perpendicular to the reaction force R (along the surface of the material) and opposite to the direction of motion with unitary vector v. Hence we can express this as,

Ff = − μR v,

although the magnitude will usually suffice.

Now, we shall further examine what is meant by the reaction force. The reaction force is defined in a direction perpendicular to the surface the object rests upon and will often have its y-component Rx equal in magnitude and opposite in direction to the weight W of the object, assuming the weight acts purely in the y-direction. See Newton’s third law of motion for more information. Thus we have,

Rx = W,

assuming the object is fully supported by the surface it sits upon.

These results can be useful when applied simultaneously. If we know the frictional force and friction coefficient, we can find the reaction force and, in turn, the object’s weight. Since we know that

W = mg

from Newton’s second law of motion, we can hence find the object’s mass and its overall acceleration.

Step 1: Draw free-body diagram. 

(only 2 forces here: normal force N and weight W = mg) 

 Step 2: Choose coordinate system. Here I chose tilted xy 

coordinates because I want the +x direction to be along 

the direction of the acceleration. 

Step 3: Write down equations 

Fx=mx*ax , Fy= my*ay 

Notice that as Theta –> 0 (track becomes horizontal), cos–> 1, sin –> 0, N –>mg, a –> 0, as expected. 

NOTICE that N is NOT equal to mg. The equation N = mg is only true in a very special situation: when the mass m is NOT accelerating and is sitting on a horizontal surface. 

Dubson’s Law: In general, N ≠ mg The statement N = mg is NOT a law. It is only true in special cases. 

  • Physics teacher: *points to free-body diagram* What's the force?
  • Half the class: "The Force is an energy field created by all living things, it surrounds us, it penetrates us, it binds the galaxy together"
  • Teacher and rest of the class: Nerds. *facepalm*
I am, and ever will be, a white socks, pocket protector, nerdy engineer – born under the second law of thermodynamics, steeped in steam tables, in love with free-body diagrams, transformed by Laplace, and propelled by compressible flow.

– Neil Armstrong

Read about Neil Armstrong in this tribute published in The Economist in 2012:

Read the post by MIT student blogger Allan K. ‘17: