Phys 207 Announcements z

Hwk3 submission deadline = 1st thing this morning

include DSC number on homework

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Today’s Agenda Ch. 4 problems on board More discussion of dynamics ÍRecap ÍThe tools we have for making & solving problems: » Ropes & Pulleys (tension) » Hooke’s Law (springs) ÍProblems: Accelerometer, inclined plane, motion in a circle 1

Review z

Discussion of dynamics. ÍNewton’s 3 Laws ÍThe Free Body Diagram ÍThe tools we have for making & solving problems: » Ropes & Pulleys (tension) »Hooke’s Law (springs)

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Page 1

Review: Pegs & Pulleys z

Used to change the direction of forces ÍAn ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: The tension is the same on both sides! massless rope F1 = -T i ideal peg or pulley

| F1 | = | F2 | F2 = T j

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Springs z

Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. ÍFX = - k x

Where x is the displacement from the relaxed position and k is the constant of proportionality. relaxed position

FX = 0 x 4

Page 2

Springs... z

Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. ÍFX = - k x

Where x is the displacement from the relaxed position and k is the constant of proportionality. relaxed position FX = -kx > 0 x

x0

x 6

Page 3

Scales: z

Springs can be calibrated to tell us the applied force. Í We can calibrate scales to read Newtons, or... ÍFishing scales usually read weight in kg or lbs.

1 lb = 4.45 N

0 2 4 6 8

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Lecture 9, Act 1 Springs z

A spring with spring constant 40 N/m has a relaxed length of 1 m. When the spring is stretched so that it is 1.5 m long, what force is exerted on a block attached to the end of the spring?

x=0

k

x=1

x=0

x = 1.5

k

M

M

(a) -20 N

(b) 60 N

(c) -60 N

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Page 4

Lecture 9, Act 1 Solution Recall Hooke’s law:

z

Í FX = - kx Where x is the displacement from equilibrium.

FX = - (40) ( .5) FX = - 20 N

(a) -20 N

(b) 60 N

(c) -60 N

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Problem: Accelerometer z

A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road, and has an acceleration a in the x direction. The string makes an angle θ with respect to the vertical (y) axis. Solve for θ in terms of a and g.

a

θ

i

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Page 5

Accelerometer... z

Draw a free body diagram for the mass: Í What are all of the forces acting? T (string tension) θ

m i

mg (gravitational force)

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Accelerometer... z

Using components (recommended):

i: FX = TX = T sin θ = ma

TX

θ T

j: FY = TY − mg = T cos θ − mg = 0

θ

TY

j

m ma

i mg

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Page 6

Accelerometer... z

Using components :

i: T sin θ = ma

TX

j: T cos θ - mg = 0 T z

Eliminate T :

θ

TY

j

m

T sin θ = ma T cos θ = mg

tan θ =

ma

a g

i mg

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Accelerometer... z

Alternative solution using vectors (elegant but not as systematic):

z

Find the total vector force FNET: T mg θ

T (string tension) θ m

FTOT

mg (gravitational force)

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Page 7

Accelerometer... z z z

Alternative solution using vectors (elegant but not as systematic): Find the total vector force FNET: Recall that FNET = ma: T (string tension) θ T mg θ m ma mg (gravitational force)

z

So tan θ =

ma a = mg g

tan θ =

a g

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tan θ =

a g

Accelerometer...

z

Let’s put in some numbers:

z

Say the car goes from 0 to 60 mph in 10 seconds: Í 60 mph = 60 x 0.45 m/s = 27 m/s. ÍAcceleration a = ∆v/∆t = 2.7 m/s2. Í So a/g = 2.7 / 9.8 = 0.28 . Í θ = arctan (a/g) = 15.6 deg a

θ

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Page 8

Problem: Inclined plane z

A block of mass m slides down a frictionless ramp that makes angle θ with respect to the horizontal. What is its acceleration a ?

m

a

θ

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Inclined plane... z

Define convenient axes parallel and perpendicular to plane: Í Acceleration a is in x direction only.

j m

a

θ

i

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Page 9

Inclined plane...

z

Consider x and y components separately: i: mg sin θ = ma. a = g sin θ

z

j: N - mg cos θ = 0.

z

N = mg cos θ

ma

j

mg sin θ N θ mg cos θ

i

mg 19

Inclined plane... z

Alternative solution using vectors:

j m N θ mg i a = g sin θ i N = mg cos θ j

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Page 10

Angles of an Inclined plane The triangles are similar, so the angles are the same!

ma = mg sin θ

N

θ mg θ

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Lecture 9, Act 2 Forces and Motion z

A block of mass M = 5.1 kg is supported on a frictionless ramp by a spring having constant k = 125 N/m. When the ramp is horizontal the equilibrium position of the mass is at x = 0. When the angle of the ramp is changed to 30o what is the new equilibrium position of the block x1? (a) x1 = 20cm

(b) x1 = 25cm

(c) x1 = 30cm

x1 =

k

k

x=0 M

? M

θ = 30o 22

Page 11

Lecture 9, Act 2 Solution z z z

Choose the x-axis to be along downward direction of ramp. FBD: The total force on the block is zero since it’s at rest. Consider x-direction: Force of gravity on block is Fx,g = Mg sinθ Force of spring on block is Fx,s = - kx1 N x1

y

F x,s

=-

kx 1

k M

x

Fx,g = Mg sinθ θ

θ

Mg 23

Lecture 9, Act 2 Solution z

Since the total force in the x-direction must be 0: x1 =

Mg sinθ − kx1 = 0 x1 =

Μg sin θ κ

5.1kg ⋅ 9.8 1m s 2 ⋅ 0.5 = 0 .2 m 125 N m x1

F x,s

y

x1 = -k

k M

x θ

F x,g

g =M

sinθ

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Page 12

Problem: Two Blocks z

Two blocks of masses m1 and m2 are placed in contact on a horizontal frictionless surface. If a force of magnitude F is applied to the box of mass m1, what is the force on the block of mass m2?

F

m1

m2

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Problem: Two Blocks z

Realize that F = (m1+ m2) a :

z

Draw FBD of block m2 and apply FNET = ma:

F / (m1+ m2) = a

F1,2 = m2 a

z

F1,2

m2

Substitute for a :

⎛ ⎞ F ⎟⎟ F1,2 = m2 ⎜⎜ ⎝ (m1 + m2 )⎠

F1,2 =

m2 F (m1 + m2) 26

Page 13

Problem: Tension and Angles z

A box is suspended from the ceiling by two ropes making an angle θ with the horizontal. What is the tension in each rope?

θ

θ m

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Problem: Tension and Angles z

Draw a FBD:

T1 T1sin θ

θ T1cos θ

T2 j

T2sin θ

θ T2cos θ

i

mg z

Since the box isn’t going anywhere, Fx,NET = 0 and Fy,NET = 0 Fx,NET = T1cos θ - T2cos θ = 0

T1 = T2

Fy,NET = T1sin θ + T2sin θ - mg = 0

T1 = T2 =

mg 2 sin θ 28

Page 14

Problem: Motion in a Circle z

A boy ties a rock of mass m to the end of a string and twirls it in the vertical plane. The distance from his hand to the rock is R. The speed of the rock at the top of its trajectory is v. ÍWhat is the tension T in the string at the top of the rock’s trajectory? v T

R

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Motion in a Circle... z z z

Draw a Free Body Diagram (pick y-direction to be down): We will use FNET = ma (surprise) y First find FNET in y direction: mg

FNET = mg +T

T

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Page 15

Motion in a Circle... FNET = mg +T v z

y

Acceleration in y direction: mg

ma = mv2 / R

T

F = ma

mg + T = mv2 / R R T = mv2 / R - mg

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Motion in a Circle... z

What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp? Íi.e. find v such that T = 0.

v

mv2

/ R = mg + T

mg

v2 / R = g

T= 0

v = Rg z

Notice that this does not depend on m.

R

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Page 16

Lecture 9, Act 3 Motion in a Circle z

A skier of mass m goes over a mogul having a radius of curvature R. How fast can she go without leaving the ground? v mg N R

Rg (a) v = mRg (b) v = m

(c) v = Rg 33

Lecture 9, Act 3 Solution z

mv2 / R = mg - N

z

For N = 0: v = Rg

v mg N R

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