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 8 th, 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. 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 T 1 and T 2 ? Note: No lever arm. Thus no torques. Two equations with two unknowns, can solve for T 1 and T 2 after some algebra.

Static Equilibrium Problem: 4 Step Process #1 – Draw the free body diagram identifying all forces and exactly where they act. Mgmg N #2 – Label the coordinate axis and axis of rotation being considered (and the sign convention) x y Axis of rotation CCW= +

Static Equilibrium Problem: 4 Step Process Mgmg N x y Axis of rotation CCW= + #3 – Write out F net = ma and  net = I  equations… F net,x = (M+m)a x = 0 (no forces in this direction) F net,y = (M+m)a y = 0 = +N – Mg – mg  net = I  = 0 = +(Mg)D 1 – (mg)D 2 #4 – Solve…

Clicker QuestionRoom Frequency BA Assume pencil of length L and mass M and with  =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.  What is the torque around the tip of the pencil at this time? A)-(N-Mg) sin(60) L B)-Mg cos(60) L/2 – N cos(60) L C)-Mg sin(60) L/2 D)-Mg cos(60) L/2 E)-Mg sin(60) L Mg N Mgcos(60) Mgsin(60) L/2  net = r F perp = -(L/2) (Mgcos(60))

Stability and Balance If 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 QuestionRoom Frequency BA Case 1Case 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 equilibrium B)Case 1 is in unstable equilibrium and Case 2 is just unstable C)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 unstable E)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 m B) 2 m C) 3 m D) 4 m E) 6 m Clicker QuestionRoom Frequency BA Ignoring any small forces holding it to the ground.

A sign with mass m s is hung from a uniform bar of mass m b 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 m s is hung from a uniform horizontal bar of mass m B 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)F wx = 0. Using Σ F x = 0, the hinge force must have a positive x-component, in order to cancel the negative x-component of the tension force. Clicker QuestionRoom Frequency BA

x y CCW= + Step #3: Static Equilibrium condition F=ma=0 and  =I  =0 Not enough information to solve for T, F wx, F wy (2 constraint equations and 3 unknowns)