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AP Physics 1
Framework, Linear Forces

How to Solve All Difficult Force Problems – Framework

Jason
14 minutes
1 formulas
10 questions
2 videos

14 minutes
1 formulas
10 questions
2 videos

Picture of Jason Kuma

Jason Kuma

Writer | Coach | Builder | Fremont, CA

Get Expert Physics Coaching

This post will help you solve any force (dynamics) problems in Physics via a 3 step framework.

Why You’re Here

In Physics, there are an endless amount of force scenarios. These include pushing objects, hanging masses, elevators, pulleys, objects on inclines, etc.

Memorizing equations for each situation would be impossible! Instead we derive equations from scratch to solve all Physics problems.

You’re here to learn, as quickly as possible, how to derive an equation and solve any linear force problem.

This can be quite difficult for some students, but the simple 4 step framework makes it easy.

The Force Framework

A framework is a general procedure to tackle a specific problem. Remember that this framework will guide you to solve problems. This is NOT a substitute for logic.

Read the problem and turn the main object into an FBD. Make sure to think about any non-obvious forces, like friction or normal forces, that maybe acting on the object
Break apart force vectors at angles into their x and y components.
Find the net force in the x and y direction separately. This can be done with real numbers or in terms of variables, if not given numbers.
Use F_net = ma! Set the net force that you found equal to ma. Rearrange and solve for the missing variable.

LRN Forces

In this video I explain the framework and how to apply it to a problem. Jump to the next section in this article if you want to see me apply it to harder problems.

PS – Force Problems

This framework above can be confusing. So in the video below I go over 3 example problems to help you better understand the process.

Picture of Jason Kuma

Jason Kuma

Writer | Coach | Builder | Fremont, CA

Programs

Units in AP Physics 1

Unit 1 – Linear Kinematics

Unit 2 – Linear Forces

Unit 3 – Circular Motion

Unit 4 – Energy

Unit 5 – Momentum

Unit 6 – Torque

Unit 7 – Oscillations

Unit 8 – Fluids

Reading Key

LRN

RE

PS

PQ

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Kinematics	Forces
$\Delta x = v_i t + \frac{1}{2} at^2$	$F = ma$
$v = v_i + at$	$F_g = \frac{G m_1m_2}{r^2}$
$a = \frac{\Delta v}{\Delta t}$	$f = \mu N$
$R = \frac{v_i^2 \sin(2\theta)}{g}$

Circular Motion	Energy
$F_c = \frac{mv^2}{r}$	$KE = \frac{1}{2} mv^2$
$a_c = \frac{v^2}{r}$	$PE = mgh$
	$KE_i + PE_i = KE_f + PE_f$

Momentum	Torque and Rotations
$p = m v$	$\tau = r \cdot F \cdot \sin(\theta)$
$J = \Delta p$	$I = \sum mr^2$
$p_i = p_f$	$L = I \cdot \omega$

Simple Harmonic Motion
$F = -k x$
$T = 2\pi \sqrt{\frac{l}{g}}$
$T = 2\pi \sqrt{\frac{m}{k}}$

Constant	Description
$g$	Acceleration due to gravity, typically $9.8 , \text{m/s}^2$ on Earth’s surface
$G$	Universal Gravitational Constant, $6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2$
$\mu_k$ and $\mu_s$	Coefficients of kinetic ( $\mu_k$ ) and static ( $\mu_s$ ) friction, dimensionless. Static friction ( $\mu_s$ ) is usually greater than kinetic friction ( $\mu_k$ ) as it resists the start of motion.
$k$	Spring constant, in $\text{N/m}$
$M_E = 5.972 \times 10^{24} , \text{kg}$	Mass of the Earth
$M_M = 7.348 \times 10^{22} , \text{kg}$	Mass of the Moon
$M_M = 1.989 \times 10^{30} , \text{kg}$	Mass of the Sun

Variable	SI Unit
$s$ (Displacement)	$\text{meters (m)}$
$v$ (Velocity)	$\text{meters per second (m/s)}$
$a$ (Acceleration)	$\text{meters per second squared (m/s}^2\text{)}$
$t$ (Time)	$\text{seconds (s)}$
$m$ (Mass)	$\text{kilograms (kg)}$

Variable	Derived SI Unit
$F$ (Force)	$\text{newtons (N)}$
$E$ , $PE$ , $KE$ (Energy, Potential Energy, Kinetic Energy)	$\text{joules (J)}$
$P$ (Power)	$\text{watts (W)}$
$p$ (Momentum)	$\text{kilogram meters per second (kgm/s)}$
$\omega$ (Angular Velocity)	$\text{radians per second (rad/s)}$
$\tau$ (Torque)	$\text{newton meters (Nm)}$
$I$ (Moment of Inertia)	$\text{kilogram meter squared (kgm}^2\text{)}$
$f$ (Frequency)	$\text{hertz (Hz)}$

General Metric Conversion Chart

Conversion Example

Example of using unit analysis: Convert 5 kilometers to millimeters.

Start with the given measurement: $\text{5 km}$
Use the conversion factors for kilometers to meters and meters to millimeters: $\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}$
Perform the multiplication: $\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}$
Simplify to get the final answer: $\boxed{5 \times 10^6 \, \text{mm}}$

Prefix	Symbol	Power of Ten	Equivalent
Pico-	p	$10^{-12}$
Nano-	n	$10^{-9}$
Micro-	µ	$10^{-6}$
Milli-	m	$10^{-3}$
Centi-	c	$10^{-2}$
Deci-	d	$10^{-1}$
(Base unit)	–	$10^{0}$
Deca- or Deka-	da	$10^{1}$
Hecto-	h	$10^{2}$
Kilo-	k	$10^{3}$
Mega-	M	$10^{6}$
Giga-	G	$10^{9}$
Tera-	T	$10^{12}$

Some answers may be slightly off by 1% depending on rounding, etc.
Answers will use different values of gravity. Some answers use 9.81 m/s², and other 10 m/s²for calculations.
Variables are sometimes written differently from class to class. For example, sometime initial velocity $v_i$ is written as $u$ ; sometimes $\Delta x$ is written as $s$ .
Bookmark questions that you can’t solve so you can come back to them later.
Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!