Announcements

• CAPA Set #5 due today (Friday) at 10 pm

• CAPA Set #6 now available, due next Friday

• Next week in Section Lab 2: Acceleration due to gravity Make sure to do the pre-lab and print materials before lab • Complete reading all of Chapter 4

• Over the weekend, read Chapter 5 Sections 5.1-5.5

m

Rock Climber hanging by ropeObject of mass m suspended by a cord.

Stationary: v = 0 and a = 0.

+y

0 = ma = Fnet

T

Fg=mg

Tension in the cord is equal to the weight

of the object.

0 = T - mgT = mg

m

Person of mass m suspended by a rope.+y

T mg Fg

Now, imagine the person is being raised at a constant velocity v > 0.

What’s the relation between T and mg?

v If v=0:

A) T = mgB) T > mgC) T < mgD) Indeterminate from information given.

v = constant, a = 0 again, as in the stationary case.

T

mg = Fg

0 ma T Fg T mg

T mg

Clicker Question Room Frequency BA

m

Object of mass m suspended by a cord.

v = constant +y

T mg Fg

Now, imagine the object is being raised with an acceleration a > 0.

What’s the relation between T and mg?

aBefore:

A) T = mgB) T > mgC) T < mgD) Indeterminate from information given.

T

-mg = Fg

0 ma T Fg T mg

T m(a g) mg

Clicker Question Room Frequency BA

http://en.wikipedia.org/wiki/Atwood_machine

The Atwood machine was invented in 1784 by Rev.

George Atwood as a laboratory experiment to verify the

mechanical laws of motion with constant acceleration.

Mm

Two objects with masses M > m suspended from a

stationary pulley.+x

+xStep #1: Choose a coordinate system.

M m+x

Odd coordinate system (curves around pulley).This choice means that the acceleration a for both masses will be the same (direction + magnitude).

Atwood Machine (Pulley)

M

m

+x

+x

Step #2: Specify the forces (free-body diagram)

Fg=-mg

+T

Fg=+Mg

-T

Accelerations a must be same for both objects,otherwise the string would stretch or break.

MM TMgMa mgTma mm

Consider the pair of suspended masses.Assume that the pulley is frictionless and

the cord does not stretch.

What is the relationship between the tension TM in the cord above larger mass

M and the tension Tm above the smaller mass?

A) TM = Tm

B) TM > Tm

C) TM < Tm

M > m

Tension along thecord is constant.

Clicker Question Room Frequency BA

TmTM

Step #3: Solve the Equations

TMgMaF MM

mgTmaF mm mM aa

MaMgT

mamgT

gmM

mMa

Two equations.Two unknowns (a, T)

Check if the answer makes sense…..

M1

M2

Frictionless table & pulley.

1)Choose coordinates2)Identify forces.3) Write down F = ma

for each object

+x

+x

+T-T

Fg = M2g

Mass 1: M1a = T

Mass 2: M2a = M2g - T(A) M 2a T

(B) M 2a T M 2g

(C) M 2a M 2g T

(D)M 2a M 2g

What is F = ma for mass M2?

Clicker Question Room Frequency BA

Mass 1: M1a = T

Mass 2: M2a = M2g - T Substitute forT in the lower

equation

M2 a = M2g –M1a

Solve for a

a M 2

M1 M 2

g

M1

M2

Frictionless table & pulley.

1)Choose coordinates2)Identify forces.3) Write down F = ma

for each object

+x

+x

+T-T

Fg = M2g

Pulley AdvantageIn a system with multiple pulleys

(for which there are various possible arrangements), we loop a single continuous rope.

There is only one tension in the rope throughout its length.

Clicker Question Room Frequency BA

What is the force (EFFORT) requires to prevent the LOAD from falling?A) 0 lbs.B) 25 lbs.C) 50 lbs.D) 100 lbs.E) 400 lbs.

Hint: Draw a free-body diagram.

4 Ropes with

Tension T pulling up

Mg

Demonstration

FrictionWhy have you been ignoring me all this time?

Force resisting the movement of two objects in physical contact past each other.

Force of Friction depends on:1. Characteristics of materials

2. Vertical force pushing the materials together