Static Equilibrium. Lecture 22. Chapter 12. Physics I Department of Physics and Applied Physics

Transcription

1 Lecture 22 Chapter 12 Physics I Static Equilibrium Course website: Lecture Capture:

2 Outline Chapter 12 The Conditions for Equilibrium Solving Statics Problems Stability and Balance

3 Exam III Info Exam III Wed Dec. 4, 9:00-9:50am, OH 150. Exam III covers Chapters 9-11 Same format as Exam II Prior Examples of Exam III posted Ch. 9: Linear Momentum (no section 10) Ch. 10: Rotational Motion (no section 10) Ch. 11: Angular Momentum; General Rotation (no sections 7-9) Exam Review Session TBA, Ball 210

4 There are three branches of Mechanics: Kinematics Motion Forces Dynamics Motion Forces Statics Motion Forces

5 The 1st Condition for Equilibrium prevents translational motion N. 2 nd law describes translational motion F ma He doesn t want to have any sliding of a ladder, i.e. a 0 F 0 x y z F 0 F 0 F 0 An object with forces acting on it, but with zero net force, is said to be in equilibrium.

6 The 2nd Condition for Equilibrium prevents rotational motion Rotational N. 2 nd law describes rotational motion: x v ext I He doesn t want to have any rotation of a ladder, i.e. 0 ext 0 There must be no net external torque around any axis (the choice of axis is arbitrary). y 0 0 z 0

7 ConcepTest 1 Static equilibrium Consider a light rod subject to the two forces of equal magnitude as shown in figure. Choose the correct statement with regard to this situation: (A) The object is in force equilibrium but not torque equilibrium. (B) The object is in torque equilibrium but not force equilibrium (C) (D) The object is in both force equilibrium and torque equilibrium The object is in neither force equilibrium nor torque equilibrium F F F ext I 0 force equilibrium torque equilibrium X r r Here, the 1st condition is satisfied but the 2 nd isn t, so there will be rotation. So, to have static situation, both conditions must be satisfied.

8 Reduce # of Equilibrium Equations For simplicity, we will restrict the applications to situations in which all the forces lie in the xy plane. 1 st condition: x F 0 F 0 F 0 2 nd condition: x y z x y z There are three resulting equations, which we will use 1) F 0 2) F 0 3) 0 y z

9 Axis of rotation for the 3 rd equation z 0 Does it matter which axis you choose for calculating torques? NO. The choice of an axis is arbitrary If an object is in translational equilibrium and the net torque is zero about one axis, then the net torque must be zero about any other axis F 0 We should be smart to choose a rotation axis to simplify problems Any axis of rotation works

10 Any axis of rotation works for the 3 rd equation (proof) The choice of an axis is arbitrary z 0

11 Concurrent/Nonconcurrent forces F F 2 1 F 3 Concurrent forces: when the lines of action of the forces intersect at a common point, there will be no rotation. So z F F F 0 F x y x F 3 Nonconcurrent forces: when the lines of action of the forces do not intersect at a common point, there will be rotation. So z F F 0 y

12 Example:traffic light Find the tension in the two wires supporting the traffic light

13 Torque due to gravity Here, we will often have objects in which there is a torque exerted by gravity. There is a simple rule how to get gravitational torque: R CM Mg MgRsin R CM CM R CM As if the whole mass were here W Mg Mg For extended objects, gravitational torque acts as if all mass were concentrated at the center of mass

14 Recall how to calculate torque Line of force (action) r 1 r 1 r 2 r 2 F 1 F 2 r F rf sin 1 r1 F1 Direction: out of the board 2 r2 F2 Direction: into the board Draw a line of force Find the perpend. distance (r ) from the axis of rotation to that line. Magnitude of torque is r F Use the right-hand rule to find its direction

15 Ladder stability A uniform ladder of mass m and length l leans at an angle θ against a frictionless wall. If the coefficient of static friction between the ladder and the ground is μ s, determine a formula for the minimum angle at which the ladder will not slip. The forces are nonconcurrent, so we need all equilibrium conditions F w z 0 F 0 F 0 x y F gy CM W mg F gx

16 Ladder stability

17 Stability and Balance Assume that we found equilibrium of a system x F 0 0 Fy 0 z What is going to happen with the system if we disturb it slightly? Is it stable or unstable?

18 Stable equilibrium Equilibrium position Disturbed system Net force returns the ball back to its equilibrium If the forces on an object are such that they tend to return it to its equilibrium position, it is said to be in stable equilibrium.

19 Unstable equilibrium Equilibrium position Disturbed system Fx 0 Fx 0 Fy z 0 0 Fy z 0 0 r mg If, however, the forces tend to move it away from its equilibrium point, it is said to be in unstable equilibrium. Torque turns the pencil away from equilibrium

20 Border between a stable/unstable equilibrium An object in stable equilibrium may become unstable if it is tipped so that its center of mass (CM) is outside its base of support. Of course, it will be stable again once it lands! inside its base of support outside its base of support

21 Stability and Balance People carrying heavy loads automatically adjust their posture so their center of mass is over their feet. This can lead to injury if the contortion is too great.

22 ConcepTest 2 Tipping Over I A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a blue dot in each case. In which case(s) does the box tip over? A) all B) 1 only C) 2 only D) 3 only E) 2 and 3 The torque due to gravity acts like all the mass of an object is concentrated at the CM. Consider the bottom right corner of the box to be a pivot point. If the box can rotate such that the CM is lowered, it will!!

23 1. Choose one object at a time, and make a free-body diagram by showing all the forces on it and where they act. 2. Choose a coordinate system and resolve forces into components. 3. Write equilibrium equations for the forces. 4. Choose any axis perpendicular to the plane of the forces and write the torque equilibrium equation. A clever choice here can simplify the problem enormously. 5. Solve. Solving Statics Problems

24 Thank you See you on Wednesday