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AP Physics 1 | Beginner Guides
Forces, Framework, Problem Solving

Unit 2, Lesson 4 of 8

Unit 2.4 | Easily Solve Any Linear Force Problems and Deriving Equation

Jason
45 minutes
4 formulas
20 questions
2 videos

45 minutes
4 formulas
20 questions
2 videos

Picture of Jason Kuma

Jason Kuma

Writer | Coach | Builder | Fremont, CA

Get Expert Physics Coaching

You might have heard that Physics involves a lot of equations. Most of these, however, will not be given.

By the end of the lesson you will be able to craft your own equations to solve the hardest of problems.

This course article contains videos that can only be accessed once enrolled in the Learn AP Physics from Scratch Course.

Unit 2 Breakdown

You are on Lesson 4 of 8:

In this lesson:

Derivation vs Memorization
Differentiate between static and non static situations
How to begin any problem – a framework for force problems
Solving single-body linear force problems

Derivation vs Memorization

A HUGE misconception is that you will be given equations. This is false.

Physics, especially AP Physics 1, is all about deriving or making your own equations.

This is what makes Physics so fun! No one is given the right equation. To solve a problem you must create your own equation as you will learn in this lesson.

It’s a game of strategy, logic, and basic algebra.

Types of Force Problems

This Unit will cover problems only involving linear forces.

This includes tension, pulleys, ramps, friction, and many a few other types of problems. Sometimes it will be a combination of them.

Static vs Non-Static Systems

To help draw your FDB, you need to determine if the system is static or non-static. Here how:

Static System	Non-Static System
Zero net force both x and y directions (aka: in equilibrium)	Net force in x and/or y directions (aka: not in equilibrium)
Not Accelerating	Accelerating in the direction of net force
Example: Sign hanging from two cables	Example: Pulling a box across a frictionless floor

Single body vs Multi-body Systems

Another thing to look for in a problem is the number of masses (objects).

Most problems will only have one object. For example a box being pushed across a floor or a person moving in an elevator. These problems, static or non-static, can be solved using the framework below, which is the focus of this lesson.

Some problems, however, will have multiple objects. For example a pulley system with masses on both ends. Or two blocks stacked upon each other each. You will learn to solve in the last lesson of the Unit (2.8 – Solving Multi Body Systems)

Problem Solving Framework – Single Body Force Problems

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

This might be confusing. So lets apply it to a simple problem below. I’ll label the steps, so you can refer back to the framework above.

Example Problem

Problem: Using a light cord, you pull a 5kg box with a force of 20N across a rough floor. The floor has a coefficient of kinetic friction of .5 and the cord makes an angle of 30 degrees to the horizontal. What is the acceleration of the box?

Apply the framework:

Forces acting on this box include, Tension (T), kinetic friction (f_k), weight (mg), and normal force (N). Draw these forces on an FBD, like the one below.

Notice the Tension force is at a 30° angle. Break up tension in x and y components and find the value of each side. Note that you can completely solve to find a numerical value OR you can leave in terms of variables (as recommended)

Find the F_net(x) by adding up all the horizontal vectors in terms of the variables. Do the same for the vertical direction. Again, I would recommend doing this in terms of variables and not numbers.

Newton’s 2nd Law (F = ma) tell us to set first set F_net(x) = ma_x and then F_net(y) = ma_y.

Begin to solve the first equation F_net(x) for a_x , as this is what the problem asks us to find.

Pause here, and notice, that we can now plug in all number EXCEPT for N (the normal force). This means we must go to our second equation (F_net(y)) to find a value for normal force.

Now that we have a value for normal force (39.1 N), go back to the first equation F_net(x) and plug in all the numbers. This is written as step #4 in the image below.

LRN – Force Framework

Still having trouble understanding the force problem solving framework? Watch the short video below where I go over an example problem and apply the framework.

Practice Single Body Force Problems

All questions involve single objects that are either static or non-static. Follow the framework above to solve each problem. Over the next few lessons, you will learn to solve more complex force questions.

Level 1: You push a 20kg box to the right with a force of 90 N across a rough floor (µ_k = .4). What is the box’s acceleration?
Level 2: A large 100kg sign in suspended from two ropes that are both at an angle of 45° to the horizontal. What is the tension in each rope?
Level 3: A 12.0 kg box is being pulled along level ground by a 81 N force that is directed at an angle of 18° above the horizontal. The coefficient of kinetic friction is 0.330. What is the acceleration of the box?
Level 4: You pull a 10kg wagon with an constant unknown force at a 25° angle above the horizontal. The wagon accelerates at a rate of .6 m/s². The coefficient of friction is µ_k= .25 between the between the tires and the road. How much force are you pulling with?
Level 5: Conceptual – Suppose two people are pushing two identical boxes across a floor with µ_k = .2. Person A pushes down at an angle of 45° while Person B pushes the box upwards at an angle of 45°. Which person will move the box further after 60 seconds?

Show Answers

1/2 m/s²
707.1 N
3.8 m/s²
22.97 N
Person A will have the greater net force, thus greater acceleration, thus furthest distance completed in 60 seconds.

PS – Single Body Linear Forces

In this video I use the force problem solving framework to solve the Level 1, 2, and 3 problems from above.

Bonus AP Level Question

A 50 kg box at rest on a rough horizontal surface. The surface has a value of µ_s = .75 and a µ_k = .3. If you pushed the box with 300 N horizontal force, how fast would the box accelerate? If you pushed the box with a horizontal force of 500 N, how fast would the box accelerate?

Show Answer

If pushed with a force of 300 N, the box would NOT accelerate (a = 0 m/s²).

If pushed with a force of 500 N, the box would overcome static friction and acclerate at a rate of a = 7 m/s².

Lesson 2.4 Recap

In this lesson we covered a framework for solving static and non-static force problems that involved only ONE object.

You practiced several questions to test your understanding and learned how to derive equations from scratch! Now you are ready to tackle questions involving all the other linear forces!

Lesson 2.5 Preview

In the next lesson ( Unit 2.5 – Tension and Elevators) we will cover the tension forces in depth, and how we can apply it to common scenarios like an elevator. We will also cover what apparent weight is and why it matters in elevators.

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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White

Blue

Orange

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Dark Mode Equation Sheet (Download)

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!