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⚙️ ap physics c: mech

  >  

🚀 Unit 2

  •  ⏱️7 min read

2.1 Newton's Laws of Motion: First and Second Law 🥏

Gerardo Rafael Bote

gerardo rafael bote

Caroline Koffke

caroline koffke

(editor)

⏱️ November 5, 2020

📅

Overview

Now that you understand motion, you ask yourself: what causes these things to move in the first place? A certain guy named Isaac Newton developed many theorems and laws that help explain how certain things move and stop. In this unit especially, you will need to learn how his three laws describe the relationship between an object's motion and the forces it interacts with. Plus, in this unit, you will get to know about how circular motion plays a role in movement. You will need a good understanding of this unit to successfully learn other units in this course.

Big Idea

  • Force Interactions: Forces characterize interactions between objects or systems.
    • After you've stopped stirring a cup of coffee or tea, why does the swirling motion continue?
    • If you apply the same amount of "push" to a car as you would to a shopping cart, why doesn't the car move at all?
    • Why will the sun set tomorrow in nearly the same place that it set today?
    • Why must you push backward to make a skateboard move forward?

Exam Impact

Unit 2 will cover approximately 17%-23% of the exam and should take around 24, 45-minute class periods to cover. The AP Classroom personal progress check has 25 multiple choice questions and 1 free response question for you to practice on.

2.1: Newton's Laws of Motion: First and Second Law 🥏

For the first topic in this unit, you will need to focus on Newton's first 2 laws. The laws state:

Newton's First Law of Motion: A body at rest will remain at rest unless an outside force acts on it, and a body in motion at a constant velocity will remain in motion in a straight line unless acted upon by a net external force.
Newton's Second Law of Motion: If an unbalanced force acts on a body, that body will experience acceleration (or deceleration).
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2010.45-NticzAZjeo6S.png?alt=media&token=1ff0050d-8c7e-4b43-a9be-0883af8a75f5

Newton's First Law is all about inertia, which is the tendency of an object to resist acceleration. Newton's Second Law is all about finding the net force, which is where Free Body Diagrams come in handy.

Now that you know the two laws, you need to properly apply them. Be careful! Topic 2.1 covers a TON of concepts, so be sure to practice with the questions below to fully understand Newton's laws.

Always, always, always consider direction and angles!

For this unit, it is very helpful to use free-body diagrams or FBDs. Each FBD will help you determine what forces are active and interacting with the object within a certain time frame. Here is an example of what a simple FBD could look like:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-5jAAixWco3cg.png?alt=media&token=dfa241c2-bfee-4c87-afe1-bf5ede03d7d7

Image from Alberta Learning

For the FBD on the right, you can see that there are 2 forces acting on the box sitting at rest: gravitational force (weight)and normal force.

Now assume that the box has a mass of m = 7.3 kg. How can you figure out how much each force is exerting on the box? All of you have to do is apply the Second Law of Motion: ΣF = ma. ΣF is read as "the sum of all forces".

TIP: Be careful in determining how ΣF is set up! You should consider that forces are also vectors, meaning that direction is important (like in kinematics). When writing out your work for FBDs, separate the forces into their horizontal and vertical components.

For the box example, we would set up ΣF as so:

ΣF_y = F_n - F_g

Then, figure out the box's acceleration. For the box, since it as at rest and there is no vertical acceleration, then we can claim that a = 0 m/s^2. Then set up the equation:

F_n - F_g = ma = (7.3)(0)

F_n - F_g = 0

F_n = F_g = mg = (7.3)(10) = 73 N

The normal force exerts an upward force of 73 N (or 73 kg•m/s^2) while the gravitational force exerts a downward force of 73 N, putting the box at equilibrium.

Friction Force

Here are equations that you should be aware of:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2010.50-SaLOWvIFLjaY.png?alt=media&token=8adab26a-8c66-40b0-a830-35836d84d3f1

The equations above deal with friction, a force that opposes the motion of one surface past another. Friction is exerted by the interacting projections on the surface of the object and the surface interacting with the object.

Friction is always:

  • Parallel to the surface
  • Opposing the direction of motion
  • Independent from the applied force

There are two types of friction:

Static friction (friction force exerted on a stationary object), is when two surfaces do not slide relative to one another.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2010.51-Lb5a5DXjC0RQ.png?alt=media&token=e65ec88c-aad2-4bc2-a006-7279a01a7f51

apply to static friction. Pay special attention to the less than/equal to sign because the force of friction will change to fit the scenario until it reaches its maximum. The equation for the maximum is especially useful for scenarios like “the maximum angle of incline at which the block will not slip".

Kinetic friction (friction force exerted on a moving object), is when two surfaces do slide relative to one another.

Only

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2010.52-Kng6k5J4fYwP.png?alt=media&token=c83d63ad-0de6-451d-aa8e-84b758812457

applies in kinetic friction.

You will also notice that weird μ(pronounced: "mew") in both equations. That μ represents the coefficient of friction of a surface. The coefficient is always constant throughout a given surface.

Additionally, it should be noted that generally μ_s > \μ_k.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-GVJzc2vwqWra.png?alt=media&token=bc32c052-aeb4-4e64-b605-7fca205e3440

Image from Dan Levine

TIP: It is important to note that the static force increase as the exerted force increases until the object breaks free from the surface. (as shown on the graph above)

*I'm still confused... what can I do to make sure that I understand Newton's First & Second Laws?

I understand where you're coming from. When I first learned this unit, I was also confused. But, here is one keyword you should take note of: PRACTICE. Look at FBDs, look at angles, and make sure everything is in its x- and y-components!

You can also use this PhET simulation to understand how forces work in one dimension. You may have to adjust the simulation's dimensions to fit your screen. The application might also take a while to load due to your device's specifications.

https://phet.colorado.edu/sims/cheerpj/forces-1d/latest/forces-1d.html?simulation=forces-1d

Simulation from PhET

Practice Questions:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-VFkZNlGki22S.png?alt=media&token=62c412af-9ec4-4a3c-86c2-a0e1514ea687

Image from Dan Levine

a) Draw a free body diagram illustrating all forces acting on the box. b) Draw a free body diagram illustrating the net force acting on the box. c) Determine the friction force exerted on the box. d) Determine the coefficient of static friction.

Answer

a) Your FBD should look something similar to this (remember to label!):

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-aRsTzaRq1CUt.png?alt=media&token=534064e7-393e-420d-9010-2ac74dfb2980

Image from Dan Levine

b) Since the wooden box remains stationary, your FBD should look something similar to this (remember to label!):

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-rgHkHfvc8fIr.png?alt=media&token=86d98b8c-1598-44d9-9f19-15a49a9c6414

Image from Dan Levine

c) Since the wooden box is at rest, ΣF_x = 0

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2011.02-Jn2BMGMYjNKJ.png?alt=media&token=01b49ead-31aa-4c3f-9b86-38f8c32f84f4

d) Since the wooden box is at rest, ΣF_y = 0.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2011.02-ttzFW35xEIP4.png?alt=media&token=b569feeb-8dde-481a-a2de-f355dcbd7eb1

⚠️Do coefficients of friction have a unit? Nope! They are unit-less and only represent the ratio of frictional force to normal force.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-eX3KyWarVh01.png?alt=media&token=8795639d-b850-4f1f-9c0e-a18b4837d3e5

Image from Dan Levine

Answer

Use calculus here! Given that you have the position formulas, take the derivative of each equation twice to get each acceleration formula (since we are dealing with net force):

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2011.04-XBGseGdXJb6B.png?alt=media&token=0f3891aa-735e-44d6-b585-b01b9398a117

Next, determine the x-direction acceleration and y-direction acceleration of the object: a_x(2) = 10 m/s^2 and a_y(2) = 18(2) = 36 m/s^2. Then, use the Pythagorean Theorem to determine the resultant acceleration: a = sqrt{10^2+36^2} = 37.363 space m/s^2. Finally, apply Newton's 2nd law to find the magnitude of the net force:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2011.06-k2HeTNCetdPm.png?alt=media&token=be352805-9a79-483b-bfde-9ee908b04010
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-oS4q6klZ2kTt.png?alt=media&token=7eff7904-31ac-440e-91fe-6638c5ddf329

Image from collegeboard.org

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-eH0YjgQKcCYP.png?alt=media&token=0d6b6ed2-aced-49a4-86fa-7bfe0bb638b8

Image from collegeboard.org

Answer

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-zxxC5o6XLrVX.png?alt=media&token=b99366c4-e2b1-4d56-bdfb-f75050a52945

Image from collegeboard.org

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-HyCncrRVQ1Sb.png?alt=media&token=a44a67ca-80fa-4ed8-b59d-f7fc7d2e01b9

Image from collegeboard.org

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-0iaw9QnsqrJs.png?alt=media&token=f8f000de-525b-402e-adec-54223ed20a49

Image from collegeboard.org

Answer

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-DVrYMmcFoXNU.png?alt=media&token=5d7a360a-efde-4d06-b6aa-537ff4989a74

Side note: When creating a graph given the values of acceleration and two blocks of mass, it is usually created to make a graph that would demonstrate a slope of g, the acceleration due to gravity.

Two teams of nine members each engage in a tug of war. Each of the first team's members has an average mass of 68 kg and exerts an average force of 1350 N horizontally. Each of the second team's members has an average mass of 73 kg and exerts an average force of 1365 N horizontally. (Taken from Lumen Learning)

a) What is the magnitude of the acceleration of the two teams? b) What is the tension in the section of rope between the teams?

Answer

a) 0.11 m/s^2

First, start off by drawing your Free Body Diagrams. I would create a FBD for each team and include: Applied Force, Force of Gravity, Normal Force, and Tension.

Then, set up two equations of the forces in the x direction based your diagrams. Ft is the force of tension, Fa is the applied force, m is mass, and a is acceleration of the system.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2011.11-1n1ZcnRKy0mc.png?alt=media&token=bb36b644-c0fd-4e82-b2fa-90bd5f7e9417

Then you can insert the information given to you by the problem.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2011.12-8v4CuOWGLp8e.png?alt=media&token=5b8c9a24-da02-4fca-8803-f46c56cce279

Since we have two equations and two unknowns, we can use substitution to find acceleration.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2011.12-V2qguUD0Fb97.png?alt=media&token=6c269c83-a46e-4f9a-9bb4-4f9c68502ee3

b) 1.2 * 10^4 N

You can use all the work you set up for part a and substitute the acceleration into one of the equations for a team, since the tension force for both teams should be equal.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2011.13-Zg4yozMOJitI.png?alt=media&token=b3cf8181-80b9-4836-a718-fc7a0c777c23

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