1. Learning Goals 1.The conditions that must be satisfied for a body or structure to be in equilibrium. 2.What is meant by the center of gravity of a body, and how it relates to the body’s stability. 3.How to solve problems that involve rigid bodies in equilibrium.

2. We learned that a particle is in equilibrium – that is, the particle does not accelerate – in an inertial frame of reference if the vector sum of all the forces acting on the particle is zero, ∑F = 0. For an extended body, the equivalent statement is that the center of mass of the body has zero acceleration if the vector sum of all external forces acting on the body is zero. This is often called the first condition for equilibrium. Where the sum includes external forces only.

3. A second condition for an extended body to be in equilibrium is that the body must have no tendency to rotate. This means that the sum of torques due to all the external forces acting on the body must be zero about any point. To be in static equilibrium, a body at rest must satisfy both conditions for equilibrium: it can have no tendency to accelerate as a whole or to start rotating.

7. Which situation satisfies both the first and second conditions for equilibrium? 1.A seagull gliding at a constant angle below the horizontal and at constant speed; 2.An automobile crankshaft turning at an increasing angular speed in the engine of a parked car; 3.A thrown baseball that does not rotate as it sails through the air.

8. In most equilibrium problems, one of the forces acting on the body is its weight. We need to be able to calculate the torque of this force. We can calculate the toque due to the body’s weight by assuming that the entire force of gravity (weight) is concentrated at a point called the center of gravity (cg). If g has the same value at all points on a body, its center of gravity is identical to its center of mass (cm).

9. We can often use symmetry considerations to locate the center of gravity of a body, just as we did for the center of mass. The center of gravity of a homogeneous shapes is at its geometric enter. For a body with a more complex shape, we can locate the center of gravity by thinking of the body as being made of symmetrical pieces. For example, we could approximate the human body as a collection of solid cylinders, with a sphere for the head. Then we can compute the coordinates of the center of gravity of the combination from the center for mass equations: Y cm = ∑m i y i / ∑m i Z cm = ∑m i z i / ∑m i

10. When a body acted on by gravity is supported or suspended at a single point, the center of gravity is always at or directly above or below the point of suspension. Otherwise, the weight would not be in equilibrium due to a net torque. Finding the center of gravity of an irregularly shaped coffee mug.

11. When a body at equilibrium is supported at several points, its center of gravity must be somewhere within the area bounded by the supports.

12. 30 kg

13. 0.25 m

14. There are two key conditions for rigid-body equilibrium: 1.The vector sum of the forces on the body must be zero: ∑F = 0 2.The sum of the torques about any point must be zero.

19. Rank the following force magnitudes in order from greatest to smallest: 1.The weight w of the sign 2.The tension in the cable 3.The vertical component of force exerted on the rod by the hinge at P.