Physics 211 Lecture 18 Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability

ACT A (static) mobile hangs as shown below. The rods are massless and have lengths as indicated. The mass of the ball at the bottom right is 1 kg. What is the total mass of the mobile? Demo: planets mobile A) 4 kg B) 5 kg C) 6 kg D) 7 kg E) 8 kg 1 m 2 m 1 kg 1 m 3 m

ACT In which of the static cases shown below is the tension in the supporting wire bigger? In both cases M is the same, and the blue strut is massless. A) Case 1 B) Case 2 C) Same M 300 L T1 Case 1 M 300 L/2 T2 Case 2

It’s the same. Why? Balancing Torques q q M L T1 d1 Case 1 M T2 d2

CheckPoint In which of the static cases shown below is the tension in the supporting wire bigger? In both cases the red strut has the same mass and length. A) Case 1 B) Case 2 C) Same 300 L/2 T2 Case 2 T1 L M Case 1

CheckPoint In which of the static cases shown below is the tension in the supporting wire bigger? In both cases the red strut has the same mass and length. A) Case 1 B) Case 2 C) Same A) Case 1 because the lever arm distance is the largest there B) The torque caused by gravity is equal to the torque caused by the tension. Since in Case 1, the lever arm is longer, the force does not need to be as long to equalize the torque caused by gravity. C) The angles are the same so the tensions are the same, the lengths do not matter.

Nice and clear; I like this explanation. Definition? Hmmm, I think I don’t know what this means. Good; I like this. Seems to me that a longer length means smaller tension It should seem VERY similar to a ladder problem. But that is the question. Location is important for torques. But the question is about the force or tension.

Homework Problem Balance forces: T1 + T2 = Mg T2 T1 = T2 = T Same distance from CM: T = Mg/2 So: CM Mg

Homework Problem 9

These are the quantities we want to find: ACM T1 M a d Mg y x

ACT What is the moment of inertia of the beam about the rotation axis shown by the blue dot? M A) B) C) d L 11

ACT The center of mass of the beam accelerates downward. Use this fact to figure out how T1 compares to weight of the beam? ACM T1 A) T1 = Mg B) T1 > Mg C) T1 < Mg M a d Mg y x 12

ACT The center of mass of the beam accelerates downward. How is this acceleration related to the angular acceleration of the beam? ACM T1 A) ACM = da B) ACM = d /a C) ACM = a / d M a d Mg y x 13

ACM = da Use ACM = da to find a Apply ACM T1 a M Apply d Mg y x Plug this into the expression for T1

After the right string is cut, the meterstick swings down to where it is vertical for an instant before it swings back up in the other direction. What is the angular speed when the meter stick is vertical? M y x d Conserve energy: T w 15

T Applying ACM = w2d Centripetal acceleration d w y x Mg 16

We will now work out the general case Another HW problem: We will now work out the general case

General Case of a Person on a Ladder Bill (mass m) is climbing a ladder (length L, mass M) that leans against a smooth wall (no friction between wall and ladder). A frictional force f between the ladder and the floor keeps it from slipping. The angle between the ladder and the wall is f. How does f depend on the angle of the ladder and Bill’s distance up the ladder? L f Bill M m y x q f

x: Fwall = f y: N = Mg + mg Balance forces: Fwall Balance torques (measured about the foot): L/2 Mg axis d mg q f y x N

This is the General Expression: Climbing further up the ladder makes it more likely to slip: Making the ladder more vertical makes it less likely to slip: M d m q f

If it’s just a ladder… Moving the bottom of the ladder farther from the wall makes it more likely to slip:

CheckPoint In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground bigger? A) Case 1 B) Case 2 C) Same Case 1 Case 2 22

CheckPoint In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground bigger? A) Case 1 B) Case 2 C) Same Case 1 Case 2 A) Because the bottom of the ladder is further away from the wall. B) The angle is steeper, which means there is more normal force and thus more friction. C) Both have same mass. 23

But what about the friction force? I don’t know what this means. okay A Force can not be equal to a torque. The frictional force will be smaller with a smaller angle between the ladder and the wall.

CheckPoint Suppose you hang one end of a beam from the ceiling by a rope and the bottom of the beam rests on a frictionless sheet of ice. The center of mass of the beam is marked with an black spot. Which of the following configurations best represents the equilibrium condition of this setup? A) B) C)

CheckPoint Which of the following configurations best represents the equilibrium condition of this setup? A) B) C) Demo If the tension has any horizontal component, the beam will accelerate in the horizontal direction. Objects tend to attempt to minimize their potential energy as much as possible. In Case C, the center of mass is lowest.

But the wire is NOT perpendicular to the beam in any of the situations. okay I don’t know what this means. Good Okay What’s a “footprint” when something is suspended? Good Very good ? ? ? What’s a “footprint” when something is suspended? When things aren’t clear, please come see me — always! That does not seem “intuitive” to me. But it is true — from analyzing the forces and torques.

Stability & Potential Energy CM Demos footprint