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PHYSICS 231 Lecture 9: Revision. Remco Zegers Walk-in hour: Thursday 11:30-13:30 Helproom. The slides about friction are in lecture 8!!. TRIGONOMETRY. SOH-CAH-TOA: sin=opposite/hypotenuse cos=adjacent/hypotenuse tan=opposite/adjacent. Pythagorean theorem:. Units.

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physics 231 lecture 9 revision

PHYSICS 231Lecture 9: Revision

Remco Zegers

Walk-in hour: Thursday 11:30-13:30

Helproom

PHY 231

slide3

TRIGONOMETRY

SOH-CAH-TOA:

sin=opposite/hypotenuse

cos=adjacent/hypotenuse

tan=opposite/adjacent

Pythagorean theorem:

PHY 231

units
Units
  • Convert all units in your problem to be in the SI system
  • When adding/subtracting two quantities check whether their units are the same.
  • If you are unsure about an equation that you want to use, perform the dimensional analysis and make sure that each part of the equation that is set equal/subtracted/added have the same dimensions
  • When 2 quantities are multiplied, their units do not have to be the same. The result will have as unit the multiplied units of the quantities being multiplied.
  • Sin, cos and tan of angles are dimensionless

PHY 231

solving a quadratic equation
Solving a quadratic equation

At2+Bt+c=0

  • Use the equation:
  • CHECK THE ANSWER
  • OR just calculate it!
  • t2+(B/A)t+C/A=0 and use (t+d)2=t2+2dt+d2
  • [t+(B/2A)]2-B2/4A2+C/A=0
  • t=-B/2A(B2/4A2-C/A)

PHY 231

slide6

Constant velocity

Constant acceleration

Constant motion

x(t)=x0+v0t+½at2

x(t)=x0

x(t)=x0+v0t

x(m)

x(m)

x(m)

t

t

t

v

m/s

v

m/s

v

m/s

v(t)=0

v(t)=v0+at

v(t)=v0

t

t

t

a

m/s2

a

m/s2

a

m/s2

a(t)=0

a(t)=0

a(t)=a0

t

t

t

PHY 231

slide7

x

v

t

time

PHY 231

slide8

A+B

Vector operations in equations

(xa+b,ya+b)

y

(xb,yb)

B

(xa,ya)

A

x

Xa=Acos()

Ya=Asin()

length/magnitude of A: (Xa2+Ya2)

PHY 231

slide9

vx=v0cos

vy=v0sin-2g=0

vx=v0cos

vy=v0sin-1g

vx=v0cos

vy=v0sin-3g

V=v0

vx=v0cos

vy=v0sin

vx=v0cos

vy=v0sin-4g

vx=v0cos

vy=v0sin-5g

Parabolic motion: decompose x and y directions

t=0

t=1

t=2

t=3

t=4

t=5

PHY 231

parabolic motion

X=X0

Y=Y0

Parabolic motion

X(t)=X0+V0cost

Y(t)=Y0+V0sint-1/2gt2

t=0

t=1

t=2

t=3

t=4

t=5

PHY 231

slide11

2D motion

  • When trying to understand the motion of an object in 2D decompose the motion into vertical and horizontal components.
  • Be sure of your coordinate system; is the motion of the object you want to study relative to another object?
  • Write down the equations of motion for each direction separately.
  • If you cannot understand the problem, draw motion diagrams for each of the directions separately.
  • Make sure you understand which quantity is unknown, and plug in the equation of motions the quantities that you know (givens). Then solve the equations.

PHY 231

slide12

Newton’s Laws

  • First Law: If the net force exerted on an object is zero the object continues in its original state of motion; if it was at rest, it remains at rest. If it was moving with a certain velocity, it will keep on moving with the same velocity.
  • Second Law: The acceleration of an object is proportional to the net force acting on it, and inversely proportional to its mass: F=ma
  • If two objects interact, the force exerted by the first object on the second is equal but opposite in direction to the force exerted by the second object on the first: F12=-F21

PHY 231

slide13

General strategy for problems with forces

  • If not given, make a drawing of the problem.
  • Put all the relevant forces in the drawing, object by object.
    • Think about the axis
    • Think about the signs
  • Decompose the forces in direction parallel to the motion and perpendicular to it.
  • Write down Newton’s first law for forces in the parallel direction and perpendicular direction.
  • Solve for the unknowns.
  • Check whether your answer makes sense.

PHY 231

slide14

horizontal

Vertical

h

x

y

vx

t

t

Vy

X(t)=x0+voxt

=0+12.7t

=34 (hit the

ground)

t=34/12.7=2.68 s

Y(t)=y0+v0yt-0.5gt2

=h-0.5gt2

=0 (hits the ground)

so: h=0.5gt2

h=35 m

A snowball is launched horizontally from the top of a building

at v=12.7 m/s. If it lands 34 m away from the bottom, how high was the building?

V0=12.7 m/s

h?

d=34m

PHY 231

slide15

blue

Blue ball:

yb(t)=y0+v0bt-0.5gt2=

=13.3-5.t-0.5gt2

red

13.3

v0r

x

x

Red ball:

yr(t)=yo+vort-0.5gt2=

=26.8t-0.5gt2

t

t

v0b

v

v

yb(t)=yr(t)

13.3-5t-0.5gt2=26.8t-0.5gt2

13.3=31.8t so t=0.42s

A red ball is thrown upward with a velocity of 26.8 m/s.

A blue ball is dropped from a 13.3 m high building with

initial downward velocity of 5.00 m/s. At what time will

the balls be at the same height.

Vo=-5.00

h=13.3m

V0=26.8

PHY 231

slide16

v

t

5.05+

2.96

5.09

In first period: v(t)=v0+at=0+1.48·5.09=7.53 m/s

In 2nd period: v(t)=v0+at=7.53-1.91·2.96=1.88 m/s

A car starts at rest and travels for 5.09 s with uniform

acceleration of +1.48 m/s2. The driver then brakes, causing

uniform acceleration of -1.91 m/s2. If the brakes are applied

for 2.96 s how fast is the car going after that?

PHY 231

slide17

Horizontal Vertical

Due to tide: 3000 N 0 N

Due to wind: 6000cos(135)=-4243 6000sin(135)=+4243

Sum: -1243 N 4243 N

N

6000N

Magnitude of resulting force:

Fsum=[(-1243)2+(4243)2]=4421 N

Direction: angle=tan-1(4243/-1243)=

1060 (calc: -730, add 1800)

F=ma so a=F/m=4421/2000=2.21 m/s2

W

E

3000N

S

A 2000 kg sailboat is pushed by the tide of the sea with a force of 3000 N to the East. Because of the wind in its sail

it feels a force of 6000 N toward to North-West direction.

What is the magnitude and direction of the acceleration?

PHY 231

slide18

TVerL

horizontal Vertical

left rope Tsin(45) Tcos(45)

right rope -Tsin(45) Tcos(45)

gravity 0 -1*9.81

Sum: 0 2Tcos(45)-9.81

TVerR

450

450

ThorL

Object is stationary, so:

2Tcos(45)-9.81=0 so, T=6.9 N

Fg

T

T

900

1kg

A mass of 1 kg is hanging from a rope as shown in the figure.

If the angle between the 2 supporting wires is 90 degrees,

what is the tension in each rope?

ThorR

PHY 231