# solving rigid-body equilibrium problems

There are just two key conditions for rigid-body equilibrium: The vector sum of the forces on the body must be zero, and the sum of the torques about any point must be zero. To keep things simple, we’ll restrict our attention to situations in which we can treat all forces as acting in a single plane, which we’ll call the xy-plane. Then we need consider only the x- and y-components of force in Eq. (11.1), and in Eq. (11.2) we need consider only the z-components of torque (perpendicular to the plane). The first and second conditions for equilibrium are the

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The challenge is to apply these simple conditions to specific problems. Problem-Solving Strategy 11.1 is very similar to the suggestions given in Section 5.1 for the equilibrium of a particle. You should compare it with Problem-Solving Strategy 10.1 (Section 10.2) for rotational dynamics problems.

Caution :Choosing the reference point for calculating torques In equilibrium problems, the choice of reference point for calculating torques in gtz is completely arbitrary. But once you make your choice, you must use the same point to calculate all the torques on a body. Choose the point so as to simplify the calculations as much as possible.

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Ans: 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 homoge-neous sphere, cube, or rectangular plate is at its geometric center. The center of gravity of a right circular cylinder or cone is on its axis of symmetry. view more..
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Ans: 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. The weight doesn’t act at a single point; it is distributed over the entire body. But we can always calculate the torque 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 view more..
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Ans: In this chapter we’ll apply the first and second conditions for equilibrium to situations in which a rigid body is at rest (no translation or rotation). Such a body is said to be in static equilibrium view more..
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Ans: There are just two key conditions for rigid-body equilibrium: The vector sum of the forces on the body must be zero, and the sum of the torques about any point must be zero. To keep things simple, we’ll restrict our attention to situations in which we can treat all forces as acting in a single plane, which we’ll call the xy-plane view more..
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Ans: The rigid body is a useful idealized model, but the stretching, squeezing, and twisting of real bodies when forces are applied are often too important to ignore. view more..
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Ans: The simplest elastic behavior to understand is the stretching of a bar, rod, or wire when its ends are pulled (Fig. 11.12a). Figure 11.14 shows an object that initially has uniform cross-sectional area A and length l0. We then apply forces of equal magnitude F# but opposite directions at the ends (this ensures that the object has no tendency to move left or right). We say that the object is in tension. view more..
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Ans: When a scuba diver plunges deep into the ocean, the water exerts nearly uniform pressure everywhere on his surface and squeezes him to a slightly smaller volume. This is a different situation from the tensile and compressive stresses and strains we have discussed. view more..
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Ans: The third kind of stress-strain situation is called shear. The ribbon in Fig. 11.12c is under shear stress: One part of the ribbon is being pushed up while an adjacent part is being pushed down, producing a deformation of the ribbon. view more..
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Ans: Hooke’s law—the proportionality of stress and strain in elastic deformations— has a limited range of validity. In the preceding section we used phrases such as “if the forces are small enough that Hooke’s law is obeyed.” Just what are the limitations of Hooke’s law? What’s more, if you pull, squeeze, or twist anything hard enough, it will bend or break view more..
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Ans: summary of equilibrium and elasticity view more..
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Ans: Fluids play a vital role in many aspects of everyday life. We drink them, breathe them, swim in them. They circulate through our bodies and control our weather. The physics of fluids is therefore crucial to our understanding of both nature and technology view more..
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Ans: A fluid is any substance that can flow and change the shape of the volume that it occupies. (By contrast, a solid tends to maintain its shape.) We use the term “fluid” for both gases and liquids. The key difference between them is that a liquid has cohesion, while a gas does not. The molecules in a liquid are close to one another, so they can exert attractive forces on each other and thus tend to stay together (that is, to cohere). That’s why a quantity of liquid maintains the same volume as it flows: If you pour 500 mL of water into a pan, the water will still occupy a volume of 500 mL. The molecules of a gas, by contrast, are separated on average by distances far larger than the size of a molecule. Hence the forces between molecules are weak, there is little or no cohesion, and a gas can easily change in volume. If you open the valve on a tank of compressed oxygen that has a volume of 500 mL, the oxygen will expand to a far greater volume. view more..
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Ans: A fluid exerts a force perpendicular to any surface in contact with it, such as a container wall or a body immersed in the fluid. This is the force that you feel pressing on your legs when you dangle them in a swimming pool. Even when a fluid as a whole is at rest, the molecules that make up the fluid are in motion; the force exerted by the fluid is due to molecules colliding with their surroundings view more..
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Ans: If the weight of the fluid can be ignored, the pressure in a fluid is the same throughout its volume. We used that approximation in our discussion of bulk stress and strain in Section 11.4. But often the fluid’s weight is not negligible, and pressure variations are important. Atmospheric pressure is less at high altitude than at sea level, which is why airliner cabins have to be pressurized. When you dive into deep water, you can feel the increased pressure on your ears. view more..
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Ans: Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. view more..
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Ans: If the pressure inside a car tire is equal to atmospheric pressure, the tire is flat. The pressure has to be greater than atmospheric to support the car, so the significant quantity is the difference between the inside and outside pressures. When we say that the pressure in a car tire is “32 pounds” (actually 32 lb>in.2 , equal to 220 kPa or 2.2 * 105 Pa), we mean that it is greater than atmospheric pressure (14.7 lb>in.2 or 1.01 * 105 Pa) by this amount. view more..
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Ans: The simplest pressure gauge is the open-tube manometer . The U-shaped tube contains a liquid of density r, often mercury or water. The left end of the tube is connected to the container where the pressure p is to be measured, and the right end is open to the atmosphere view more..

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