About Midterm Exam 1• When and where

– Thurs Feb. 17th 5:45-7:00 pm– Rooms: See course webpage. Be sure report to your TA’s room – Your TA will give a review during the discussion session next week.

• Format– Closed book, 20 multiple-choices questions (consult with practice exam)– 1page 8x11 formula sheet allowed, must be self prepared, no photo

copying/download-printing of solutions, lecture slides, etc.– Bring a calculator (but no computer). Only basic calculation functionality can be used. Bring a 2B pencil for Scantron.– Fill in your ID and section # !

• Special requests: – Email me at than@hep.wisc.edu, w/ valid excuse for approval.– One alternative exam: 3:30pm – 4:45pm, Thurs Feb.17, in our 201 lab rm.

Lect. 8, Chapter 4: Newton’s LawsToday: The 3rd Law; and applications

Important technique:Free-Body Diagrams

Free-body diagrams are diagrams of the forces on an object.

First, isolate the object in question.

Then, identify the individual forces on it.

Forces acting on sled:

1. The gravitational force on the sled-rope

2. The contact force exerted by the ice on the runners. (Without friction, the contact force is directed normal to the ice.)

3. The contact force exerted by the dog on the rope.

(Since the sled remains on the ice, the y-components of the force sum to zero.)

Example: Hanging a Picture

Free-body diagram for the picture

Picture is not accelerating Forces on picture sum to zero.

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x component: T1 (30cos °) - T2 (60cos °) =0 ⇒ T2 =T1(30cos °)(60cos °)

y component: T1 (30sin °)+T2 (60sin °) -mg=0

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⇒ T1 =mg

sin30°+ cos30°cos60°

sin60°= mg

12+ 3 / 2

1/ 2

⎛

⎝ ⎜

⎞

⎠ ⎟32

=mg2

; T2 =T1(30cos °)(60cos °)

= 3T1

Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is

a) larger than

b) identical to

c) less than

the downward weight W of the person.

Person is accelerating upwards - net upwards force is non zero

N – mg = ma mg

N

Question 2

Newton’s Third Law For every action, there is an equal and opposite reaction.

• Finger pushes on box • Ffingerbox = force exerted on box by finger

Ffingerbox

Fboxfinger • Box pushes on finger• Fboxfinger = force exerted on finger by box

• Third Law: Fboxfinger = - Ffingerbox

Newton’s Third Law… FFA ,B = - FFB ,A. is true for all types of forces

Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the one it exerts – either stationary or in motion.

FFw,m FFm,w

FFf,m

FFm,f

Gravity Fm,f = -mg

System of Interest

f - All forces opposing the motion

System 1: Acceleration of the professor and the cart

System 2: Force the professor exerts on the cart

Newton’s second law, combined with third law, applies to every system of interest.

The Horse Before the Cart

Free-body diagram for the cart:

Example: Forces on an object on an incline

Coordinates:Most conveniently,

Parallel & perpendicular (normal)

Normal (no motion): Fn = Fgy = mg cosθ

Parallel (moving): Fgx – Ffrict = mg sinθ – Ffrict = ma

04/21/23 Physics 103, Fall 2009, U.Wisconsin 10

Summary:• Newton’s First Law:

The motion (velocity) of an object does not change unless it is acted on by a net force

• Newton’s Second Law:Fnet = ma

• Newton’s Third Law:For every action, there is an equal and opposite reaction.

Fa,b = -Fb,a

When considering action reaction pairs it’s important to identify which forces act on a given object.

Solving Problems Identify force using Free Body Diagram

This is the most important step! Set up axes and origin

x and y Write Fnet=ma for each axis (components of forces) Calculate acceleration components Setup kinematic equations Solve!

Strong suggestion:» work problem algebraically (using symbols)» plug in numbers only at the end

Example: Pulley (Atwood Machine)

a am1=m

m2=2m

With 2 free-body diagrams, set up 2 equations:

Moving together, acceleration a is same for both blocks.

(1) T – m1g = m1a(2) m2g – T = m2a

(1) + (2) a = (m2-m1)g/(m2+m1),Thus, T = 2m1m2g/(m2+m1)

Note:(1) Sige of a matters; T is only a magnitude.(2) If not frictionless, then T not the same.

TT

a

T

W2

a

T

W1

Building a Space Station

You are an astronaut constructing a space station, and you push on a box of mass m1 with force FA1

The box is in direct contact with a second box of mass m2 .

(a) What is the acceleration of the boxes?

(b) What is the magnitude of the force each box exerts on the other?

(c) Generalization: What if for n boxes ?

(1) FA1 - F21 = m1a (2) F12 = m2a

a = FA1 /(m1 + m2), as expected

FA1 = (m1 + m2) a F12 = m2a = F21

FA1 = (m1 + m2+ … mn) a; F2 = (m2+ … mn) a; F3 = (m3+ … mn) a … … Fn = mn a