Physics 1A€¦ · investigate the forces of nature, and then from these forces to demonstrate the...

Physics 1A

Lecture 4B

• Natural motion: all things move according to their weight; heavy things go down, light things go up

• Heavy things fall faster than light things

• All things naturally come to rest

• A force is required to cause a violent motion (e.g., throwing a ball)

Aristotle (384-322 BC; a student of Plato and teacher of Alexander the Great)

Aristotelian Thinking

Many contributors through the years...

Archimedes (287-212 BC): understood density and buoyancy

ibn Sina (AD 980-1037): developed concept of momentum

Mozi (470-391 BC): concept of inertia

“The cessation of motion is due to the opposing force...If there is no opposing force...the motion will never stop. This is as true as that an ox is not a horse.”

墨翟

Kanada (2nd/6th cent. BC?): broke matter

down into atoms

Robert Hooke (AD 1635-1703): theory of elasticity, spring motion, microscope

Margaret Cavendish (AD 1623-1673): wrote 6 books arguing against Aristotelian motion

Newtonian thinking

Newton merged all of this work into three simple laws:

Sir Isaac Newton (AD 1642–1727)

1. Objects tend to remain at constant velocity unless acted on by a force

2. A force must be applied to change an object’s velocity; the acceleration depends on the mass and force applied

3. Every force has an equal and opposite counter force

5 A good read - if you like Latin

“…the whole burden of philosophy seems to consist in this — from the phenomena of motions to investigate the forces of nature, and then from these forces to demonstrate the other phenomena.”

- Newton, Principia

(1686)

• Newton synthesized centuries of work into 3 simple laws (patterns of nature)

– objects move with constant velocity (magnitude and direction) unless acted upon by a force

– acceleration of object is proportional to force applied divided by mass

– forces act in equal and opposite pairs

• Mass is the measure of inertia

• Mass is different from weight (which is mass times gravitational acceleration)

Review of Last Lecture

Newton’s Three Laws

All forces act in action–reaction pairs:

“To any action there is always an

opposite and equal reaction; in other

words, the action of two bodies upon

each other are always equal and acting

opposite in direction.”

3.

Interaction pairs

for contact forces, they occur at point of contact

Support force from table acts on block

Equal and opposite force from block acts on table

Note: Normal forces always act perpendicular to surface! N

N

Interaction pairs

W

W

Equal and opposite force of block pulling on Earth!

Earth’s gravitational pull on block (weight)

Support force from table acts on block

Equal and opposite force from block acts on table

N

N

Note: gravitational forces always act on center of mass

Six Steps to Solving Force Problems

1. Set up the problem – draw a rough figure, imagine the outcome, consider constraints on the problem; write down the quantities you are provided and the quantity you need to determine

2. Draw free-body diagrams – indicate the forces on all relevant objects, including contact forces; be sure to identify all interaction force pairs

3. Decide on an inertial coordinate system for the problem (e.g., aligned with force vectors)

4. Write down equations of motion - a = ΣF/m (= 0 for static case) - for each relevant object along each axis

5. Solve for unknowns and desired quantity(ies)

6. Check your answer (dimensions and extreme cases)

Example 1: Block on a Table at Rest

Question: what is the normal force of the table acting on the block?

m = 5.0 kg

Step 1: Conceptualize: write down what you know, what you want

m = 5.0 kg

block mass = 5.0 kg v(t) = v0 = 0 m/s => position is constant (static problem) on Earth: gravity acts downward so block has weight W = mg = (5.0 kg)(9.8 m/s2) = 49 N (down)

want: normal force (force of table on block)

Expectation: the mass is just going to sit there => this is a static force problem

Step 2: Draw free body diagrams (force vectors)

contact

W = mg = 49 N Force from block to table

Force from table to block

N

N

gravitational force of block on Earth = 4.9 N

weight of table

gravitational force of table on Earth

etc...

Note: Normal forces always act perpendicular to surface!

Step 2: Draw free body diagrams (force vectors)

contact

W = mg = 49 N

Force from table to block

N

Focus only on forces acting on block only

Step 3: Decide on an inertial coordinate system

W

N

All forces act in a line => 1D problem

Step 4: Write down equations of motion (ΣF = 0 for static) for each axis

W

N

y: ΣF = N – W = 0 => N = W = mg

= (5.0 kg)(9.8 m/s2) = 4.9 N

Step 5: Solve for unknowns

W

N

y: ΣF = N – W = 0 => N = W = mg

= (5.0 kg)(9.8 m/s2) = 49 N

Step 6: Check that solution makes sense

W

N

Correct dimensions and units

magnitude of force seems reasonable

direction and sign are OK

30º

What is the friction force that prevents the block from sliding?

Example 2: Inclined Plane (static)

Steps: 1. conceptualize 2. draw forces 3. choose coordinates

4. ΣF = 0 5. Solve for unknowns 6. Check solution

30º

know: m = 10 kg angle = 30º gravity: W = (10 kg)(9.8 m/s2) = 98 N v = v0 = 0 m/s => position constant (static problem)

want: friction force holding up block

Step 1: Write down what you know, what you want

30º

W

N Ff

N Ff

Wplane

interaction pairs

Good to draw all forces on all interacting objects In this case we do not care about the motion of the plane

Step 2: Draw forces on all relevant objects

30º

Step 2: Draw forces on all relevant objects

W

N Ff

Good to draw all forces on all interacting objects In this case we do not care about the motion of the plane

30º

Step 3: Choose an inertial coordinate system

N

W

Ff

30º

Step 4: Write down ΣF = 0 for all axes

W cos 30º W sin 30º

x: W sin 30º - Ff = 0 y: N – W cos 30º = 0

N Ff

30º

Step 5: Solve for unknowns

W cos 30º W sin 30º

x: W sin 30º - Ff = 0 y: N – W cos 30º = 0

Ff = W sin 30º = mg sin 30º = (10 kg)(9.8 m/s2)(0.5) = 49 N

N Ff

30º

Step 6: Check your solution

W cos 30º W sin 30º

right direction? ✔ right dimensions/units? ✔ answer makes sense? ✔

N Ff

30º

Step 6: Check your solution

W cos 30º W sin 30º

right direction? ✔ right dimensions/units? ✔ answer makes sense? ✔

Also check limits: Ff = W sin 30º

if angle = 0º, there would be no friction force ✔

N Ff

In review

• Six steps to solving force problems:

– conceptualize, predict, and write down knowns and desired quantities

– draw free body (force) diagrams for each object

– choose a coordinate system

– write down equations of motion (a = ΣF/m) and constraint equations

– solve for unknowns and desired values

– check your answer!

Interaction Pair: Mutual gravitational attraction

Therefore, while magnitude of the forces on each body is the same, the acceleration of the smaller mass is much higher than of the larger mass

• Third law

• Second law

r

Interaction Pair: Mutual gravitational attraction

Newton’s law of universal gravitation

r

Interaction Pair: Mutual gravitational attraction

Newton’s law of universal gravitation

• Definitions of weight and g: let r=R

For Next Time (FNT)

• Start Reading Chapter 5

• Finish the homework for Chapter 4