# Chapter 12 – Static equilibrium

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Chapter 12 – Static equilibrium
PHY 113 C General Physics I 11 AM-12:15 PM MWF Olin 101 Plan for Lecture 14: Chapter 12 – Static equilibrium Balancing forces and torques; stability Center of gravity Will discuss elasticity in Lecture 15 (Chapter 15) Problems 1.1,1.6,1.10,1.11 10/14/2013 PHY 113 C Fall Lecture 14

PHY 113 C Fall Lecture 14 10/14/2013

Newton’s law of gravitation: Earth’s gravity:
Summary of gravity: Newton’s law of gravitation: Earth’s gravity: Stable circular orbits of gravitational attracted objects: RE m RES F a v Msat 10/14/2013 PHY 113 C Fall Lecture 14

From Webassign Assignment #12:
When a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration due to the Earth's gravitation? m/s2 towards earth m r=4.4RE 10/14/2013 PHY 113 C Fall Lecture 14

From Webassign Assignment #12:
An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.79 m/s2. Determine the orbital period of the satellite. r 10/14/2013 PHY 113 C Fall Lecture 14

From Webassign Assignment #12:
How much work is done by the Moon's gravitational field as a 1090 kg meteor comes in from outer space and impacts on the Moon's surface? RM iclicker question W>0 W<0 10/14/2013 PHY 113 C Fall Lecture 14

From Webassign Assignment #12:
A space probe is fired as a projectile from the Earth's surface with an initial speed of m/s. What will its speed be when it is very far from the Earth? Ignore atmospheric friction and the rotation of the Earth. vi 10/14/2013 PHY 113 C Fall Lecture 14

From Webassign Assignment #12:
Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is v = 190 km/s and the orbital period of each is 10.7 days. Find the mass M of each star. (For comparison, the mass of our Sun is kg.) 10/14/2013 PHY 113 C Fall Lecture 14

From Webassign Assignment #12:
Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is v = 190 km/s and the orbital period of each is 10.7 days. Find the mass M of each star. (For comparison, the mass of our Sun is kg.) iclicker exercise: Who might pose a question like this? A mean professor. A puzzle master. An observational astronomer. 10/14/2013 PHY 113 C Fall Lecture 14

Meanwhile – back on the surface of the Earth:
Conditions for stable equilibrium 10/14/2013 PHY 113 C Fall Lecture 14

Stability of “rigid bodies”
N mig 10/14/2013 PHY 113 C Fall Lecture 14

Center-of-mass Torque on an extended object due to gravity (near surface of the earth) is the same as the torque on a point mass M located at the center of mass. mi ri rCM 10/14/2013 PHY 113 C Fall Lecture 14

Notion of equilibrium:
Notion of stability: T- mg cos q = 0 -mg sin q = -maq F=ma  r q T t=I a  r mg sin q = mr2 a = mraq mg(-j) Example of stable equilibrium for q » 0. 10/14/2013 PHY 113 C Fall Lecture 14

Unstable equilibrium:
Support above CM: Support below CM: r q T mg(-j) 10/14/2013 PHY 113 C Fall Lecture 14

Nik Wallenda walking on high wire across Grand Canyon
10/14/2013 PHY 113 C Fall Lecture 14

Analysis of stability:
10/14/2013 PHY 113 C Fall Lecture 14

** X 10/14/2013 PHY 113 C Fall Lecture 14

10/14/2013 PHY 113 C Fall Lecture 14

RCM Fg1 ** X mg 10/14/2013 PHY 113 C Fall Lecture 14

F1 and F2 are both up as shown. F1 is up but F2 is down.
iclicker question: F2 F1 L/3 mg Mg L Consider the above drawing of the two supports for a uniform plank which has a total weight Mg and has a weight mg at its end. What can you say about F1 and F2? F1 and F2 are both up as shown. F1 is up but F2 is down. F1 is down but F2 is up. 10/14/2013 PHY 113 C Fall Lecture 14

F2 F1 ** X L/3 mg Mg L 10/14/2013 PHY 113 C Fall Lecture 14

The fact that we found F1<0 means:
iclicker question: The fact that we found F1<0 means: We set up the problem incorrectly The analysis is correct, but the direction of F1 is opposite to the arrow Physics makes no sense iclicker question: What would happen if we analyzed this problem by placing the pivot point at F1 ?: The answer would be the same. The answer would be different. Physics makes no sense 10/14/2013 PHY 113 C Fall Lecture 14

T Mg mg ** X 10/14/2013 PHY 113 C Fall Lecture 14

** X d 10/14/2013 PHY 113 C Fall Lecture 14

Mg = 120 N mg = 98 N Fwall T < 110 N X mg Mg N ** T 10/14/2013
PHY 113 C Fall Lecture 14

A ladder of weight Mg and of length L is supported by the ground with static friction force f and by a frictionless wall as shown. The firefighter has weight mg and is half-way up the ladder. Find the force that the ladder exerts on the wall. ** X x q 10/14/2013 PHY 113 C Fall Lecture 14