Announcements• CAPA #11 due this Friday at 10 pm

• Reading: Chapter 9

• Section – this week Lab #4, next week Lab #5 (no prelab)

• Midterm Exam #3 on Tuesday November 8th, 2011 details given on course web page “exam info” practice exam and solutions on CULearn formula sheet and info. posted on web page

• Fraction of all clicker questions answered posted on CULearn. Email me with your clicker ID, name, student ID if you believe it is incorrect.

Static Equilibrium

A mass m is hanging (statically) from two strings. The mass m, and the angles α and β are known.

What are the tensions T1 and T2?Note: No lever arm. Thus no torques.

Two equations with two unknowns, can solve for T1 and T2 after some algebra.

Static Equilibrium Problem: 4 Step Process

#1 – Draw the free body diagram identifying all forces and exactly where they act.

Mg mg

N

#2 – Label the coordinate axis and axis of rotation being considered (and the sign convention)

x

y

Axis of rotationCCW= +

Static Equilibrium Problem: 4 Step Process

Mg mg

Nx

y

Axis of rotationCCW= +

#3 – Write out Fnet = ma and tnet = Ia equations…

Fnet,x = (M+m)ax = 0 (no forces in this direction)Fnet,y = (M+m)ay = 0 = +N – Mg – mg tnet = Ia = 0 = +(Mg)D1 – (mg)D2

#4 – Solve…

Clicker Question Room Frequency BA

Assume pencil of length L and mass M and with q=60 degrees.

We treat the force of gravity (Mg) as if it acts only on the center-of-mass

position of the object.

Approximately in the middle of the pencil.

q

What is the torque around the tip of the pencil at this time?A) -(N-Mg) sin(60) LB) -Mg cos(60) L/2 – N cos(60) LC) -Mg sin(60) L/2D) -Mg cos(60) L/2E) -Mg sin(60) L

MgN

Mgcos(60)

Mgsin(60)

L/2

tnet = r Fperp = -(L/2) (Mgcos(60))

Stability and BalanceIf only the gravitational force and the Normal force are acting (as shown), if the center-of-mass is not directly over

the point of contact, the system is unstable (will fall over).

System is in static equilibrium.Called neutral equilibrium.

System is in static equilibrium.Called unstable equilibrium (slightest change and it falls)

Clicker Question Room Frequency BA

Case 1 Case 2

Which of the following are correct (assume in both cases the sphere is at rest at this moment)?

A) Case 1 and Case 2 are in neutral equilibriumB) Case 1 is in unstable equilibrium and Case 2 is just unstableC) Case 1 is in neutral equilibrium and Case 2 is in neutral eq.D) Case 1 is in neutral equilibrium and Case 2 is just unstableE) None of the above

Tipping Point

If force of gravity applied at the center-of-gravity (CG) points outside the point of balance,

the object will fall (unstable).

U.S. Consumer Product Safety Commission finds approximately 2.18 deaths per year from vending machine tipovers.

7 m

55 m

4.0 m

CM

The leaning tower of Pisa is 55 m high and leans 4 m off vertical at the top. Its base is 7 m wide.

How much further could it lean horizontally (at the top) before it would fall over?

A) 1.5 mB) 2 mC) 3 mD) 4 mE) 6 m

Clicker Question Room Frequency BA

Ignoring any small forces holding it to the ground.

A sign with mass ms is hung from a uniform bar of mass mb and length L.

The sign is suspended ¾ of the way from the pivot.

The sign is held up with a cable at an angle θ.

How strong a cable is required (i.e. what is tension T)?

Step #1: Force Diagram

Step #2: Coordinate System

x

y

CCW= +

A sign of mass ms is hung from a uniform horizontal bar of mass mB as shown. What is the sign of the x-component of the force exerted on the bar by the wall? A) Positive B) Negative C) Fwx = 0.

Using Σ Fx = 0, the hinge force must have a positive x-component, in order to cancel the negative x-component of the tension force.

Clicker Question Room Frequency BA

x

y

CCW= +

Step #3: Static Equilibrium condition F=ma=0 and t=Ia=0

0, xxnet maF qcosTFwx

0, yynet maF gmgmTF sBwy qsin

Not enough information to solve for T, Fwx, Fwy (2 constraint equations and 3 unknowns)

0 at Inet ))(sin()4/3)(()2/)(( LTLgmLgm sB q

0 at Inet ))(sin()4/3)(()2/)(( LTLgmLgm sB q

qsin4)32( gmm

T sB

0, xxnet maF qcosTFwx

0, yynet maF gmgmTF sBwy qsin

qqq

sincos

4)32(

cosgmm

TF sBwx

4)32(

)(sin)(gmm

gmmTgmmF sBsBsBwy

q