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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