Monday, March 27, 2017

Newton's Second Law

Newton's Second Law is one of the most famous of all things in the realm of physics. It is fundamental to so many aspects of physics, architecture, and engineering that it is hard to find anything that equally ranks in prominence.

Stated with words, Newton's Second Law says that if an object accelerates (if it changes magnitude or direction of its velocity), the force required is equal to the mass of the object times the rate of acceleration. 

So, to state that in less fancy terms… Say something is being pushed. If you push something, it's going to change velocity. (That could mean starting or stopping.) So, if you recall that changing velocity is the same as accelerating, then when you push something it accelerates. The harder you push, the faster it accelerates. The heavier it is, the slower it accelerates. 

Yet again, logic and common since prove true! If you just stop and think about it, you already understand this.

Just for fun, let's reprise…

If something has a lot of mass, it will change velocity less under a given force than if the same force is applied to something with less mass. The stronger the force, the faster an object will accelerate.

Naturally, since this is physics, there is a formula that models the relationship between mass, force, and acceleration. Where F is force, a is acceleration, and m is mass, then:

F = ma

This leads to a couple of derived formulas, just in case finding m or a is desired.

To find F:

F = ma

To find m:

m = F/a

To find a:

a = F/m


Find the force needed to to cause a 5 kg object to accelerate at a rate of 20 (m/s)/s to the left.

F = find it
m = 5 kg
a = 20 (m/s)/s Left

F = ma
F = 5•20
F = 100 N Left


If a 20 N force to the Right is applied to an object and causes it to accelerate at a rate of 4 (m/s)/s, what is the mass of the object?

F = 20 N Right
m = find it
a = 4 (m/s)/s

F = ma
20 = 4•m
20/4 = m
5 kg = m


m = F/a
m = 20/4
m = 5 kg


If a 15 N force to the Right is applied to a 3 kg object, what will the rate of acceleration be?

F = 15 N Right
m = 3 kg
a = find it

F = ma
15 = 3•a
15/3 = a
5 (m/s)/s = a


a = F/m
a = 15/3
a = 5 (m/s)/s 


Using Newton's Second Law effectively allows demonstrating the relationship between mass, acceleration, and force. It follows logic and common sense completely, and the math requires only picking out the numbers, plugging them into the equation and using multiplication and division to solve for the missing variables.

Sunday, March 26, 2017

Units, Variables, and Formulas

When physics formulas are considered, there are many letters to think about. The formulas, themselves, are a string of variables represented by letters. Each letter represents some quantity, and the formula shows how the quantities are related to each other.

In some formulas, the displacement is considered. Displacement (and also distance) is represented by the letter d. However, many times, it is necessary to look at multiple displacements in the same situation, subscripts are used for initial and final displacements: di and df. Often, for convenience, the subscript is displayed in the same size as the variable resulting in di and df.

In other formulas, the rate that displacement changes (velocity) is considered. Velocity is represented by the letter v. However, many times, it is necessary to look at multiple velocities in the same situation, subscripts are used for initial and final displacements: vi and vf. As with displacement, for convenience, the subscript is displayed in the same size as the variable resulting in vi and vf

When formulas consider the rate at which velocity changes (acceleration), the letter a is used. 

When working with acceleration, it is sometimes necessary to look at an average velocity, which is represented as v(ave) or just v(ave). Graphically, it can be written as a v with a line on top.

Elapsed time is represented by the letter t. (It is sometimes necessary to find elapsed time by taking the difference between two clock readings, which are called—and which are confusing—initial time and final time, ti and tf).

Formulas using force will represent for with an F (capital). Working with forces often requires considering multiple forces all at once. Thus, net force becomes Fnet and the component forces get subscripted with numbers such as F1 or F2. As in other cases, for typing convenience, the subscript is sometimes rendered in the same size font as the variable.

One more variable that relates to force and acceleration is mass, and it is represented by an m. If necessary to designate different masses in the same situation, subscripts are used to designate which is which. 

The following formulas give examples of how the variables might appear in equations.

df = di + vit + 1/2at2

vf = vi + at

F = ma

v(ave) = (vi + vf)/2

One challenge to solving physics problems is working the given information into the equations. Paying attention to the units is the key to doing this successfully. The following guide will help identify where quantities with specific units should go.

Distances — df, di, d — units will be something that measures distance, such as meters (m), miles (mi), feet (ft), inches (in), centimeters (cm), etc.
3 ft
2 m
34 cm

Time — t — units will be something that measures time, such as hours (hr), minutes (min), or seconds (s).
3 s
4 min
65 hr

Velocity — vf, vi, v — units will be some distance unit divided by some time unit, as shown in the examples.
8 m/s
398 cm/s
4 mph
67 MPH
11 mi/s
39 ft/min

Acceleration — a — will be some velocity unit divided by some time unit, as shown in the examples.
3 (m/s)/s
4 MPH/s
38 (m/s)/s

Force — F, Fnet, F1, F2 — will be in newtons, a derived unit of (kg•(m/s))/s. Force should also have a direction given.
3 N Up
2 N Left

Mass — m — units will be a mass unit, such as grams (g or gm), kilograms (kg) or slugs (never mind!)
3 gm
5 kg