Physics Notes, Unit 4, Free Particle, Quantitative Problems

Includes some problems from Worksheet 4 & the Review Guide

A car parked on a hill

A car parked on a hill

• Gravity always points

toward the center of the

earth (down).

• The normal force is

perpendicular to the

road/tire surface and the

friction force is parallel to

the road/tire surface.

• Friction exerts a force up

the hill to resist the

tendency of the car to

slide down the hill due to

gravity.

A car parked on a hill

• Gravity always points

toward the center of the

earth (down).

• The normal force is

perpendicular to the

road/tire surface and the

friction force is parallel to

the road/tire surface.

• Friction exerts a force up

the hill to resist the

tendency of the car to

slide down the hill due to

gravity.

• Since the car is motionless, the forces must be balanced along each coordinate axis.

• Gravity isn't along either coordinate axis, but we can represent gravity with two component vectors.

• Fg parallel is how much of the gravitational force tends to pull the car along the slope

• and Fg perpendicular is how much of the gravitational force tends to pull the car to the road.

• For the forces to be balanced, Fg parallel must be equal in size to Ffriction and Fg perpendicular must be equal in size to Fnormal

A car parked on a hill

FN

Fg

Ff

Fpush

Fpush y (downward)

component

Fpush x (horizontal) component

These two diagrams

show the same thing

even though we moved

the upward component

of tension.

Tension force with x & y components

All forces are balanced

IB Physics Bell Work Thursday, March 26 1. Draw a force diagram for a car parked on a hill. List all forces. Draw all pictures & diagrams. a. Gravity always points to the center of the Earth. b. Gravity makes the car roll down the hill (parallel to the road). c. Gravity pulls the car to the road (perpendicular to the road). d. Friction opposes gravity parallel to the road e. The normal force supports the car perpendicular to the road.

a

b

c

d

e

surface

IB Physics Bell Work Thursday, March 26 2. Draw a force diagram for a car parked on a hill. List all forces. a. Gravity always points to the center of the Earth. b. Gravity makes the car roll down the hill (parallel to the road). c. Gravity pulls the car to the road (perpendicular to the road). d. Friction opposes gravity parallel to the road

e. The normal force supports the car perpendicular to the road.

a

b

c

d

a

IB Physics, Bell Work, Tuesday, April 7

hyp

oppsin

cos adj

hyp

tan opp

adj

𝑨 • 𝐬𝐢𝐧𝜽 = 𝑨𝒚

oppsin* hyp

adjc os* hyp

See Trigonometry Reading

𝑨 • 𝒄𝒐𝒔𝜽 = 𝑨𝒙

---------------------------------Copy on answer side -------------------------------

1. Copy on question side:

Review Sheet a. choose the simplest coordinate axis for analysis: horizontal-vertical or parallel-perpendicular. b. draw and label a force diagram for the object c. break forces not aligned with your coordinate axis into components using trigonometry. d. qualitatively use marks on the vectors to indicate equality and inequality

Fg-x

Fg-y

In this case the opposite is || to x axis So it’s the x component

Review Sheet #5

e. write equations for the vector equality marks to quantitatively calculate force values f. recognize that balanced forces always result in constant velocity (including v = 0) and unbalanced forces always cause an acceleration in the same direction as the Fnet (A child pulls a 12 kg wagon up a hill at 0.6 m/s.) a. What is force of gravity (weight) on the wagon? Fg = m•g = (12kg)10N/kg = 120 N b. What is the force the wagon exerts on the child?

FT, c, w = FT, w, c (Newton’s 3rd Law) -Ff + FT = 0 Ff = Fg ||

Fg || = Fg • sin 30°

c. What is the normal force on the wagon?

FN + (-Fg ┴ ) = 0 Fg ┴ = Fg • cos 30° = FN

FT = Ff

FN = Fg ┴

= 120 N • sin 30° = 60N

= 120 N • cos30° = 104N = FN

= FT

Unit 4, Worksheet 4, #2 A man pulls a 50 kg box at constant speed across the floor. He applies a 200 N force at an angle of 30° 1. Set your axis, draw a force diagram, label equal forces with equality marks, show component forces for a force. 2. Write an equation for all the forces and components of force on the box in

the vertical direction. 3. Write an equation for all the forces and components of force on the box in

the horizontal direction.

Fg = (FT • sin30°) + FN

Ffk = FT • cos 30° 4. Calculate the size of the normal force on the box.

(FT • sin30°) + FN = Fg (200N • sin30° ) + FN = 500 N 100N + FN = 500 N FN = 500N −100N

5. Calculate the size of the frictional force opposing the motion of the box. Ffk = FT • cos30° Ffk = 200N cos30° = 170N

FN = 400N

W= mg = 50kg 10 N/kg = 500N

W= Fg= 500N

Unit 4, Worksheet 4, #5

A 950 kg car is driven up a hill at constant velocity of 7 m/s, where 1200 N of friction and drag oppose its motion. 1. Draw a force diagram for the car. 2. What is the weight of the car? Fg = mg = 950kg(10 ) = 9500N 3. Calculate the normal force on the car. FN = Fg-y Fg-y = Fg •cos22° FN = (9500N) cos22° = 8800N 4. Calculate the force on the car that allows it to go up the hill. (Ffs) + Fg-x = Fd ||

(1200N) + sin22°(9500N) = Fd

Fd = 1200N + 3600N = 4800N

F d

Physics, Bell Work, Monday, Dec 1 a. draw and label a force diagram for the object b. choose the simplest coordinate axis for analysis: horizontal-vertical or parallel-perpendicular. c. break forces not aligned with your coordinate axis into components using trigonometry. d. qualitatively use marks on the vectors to indicate equality and inequality

Ff

Worksheet 3, #2