# Center of gravity

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
(abbreviated “cg”). The acceleration due to gravity decreases with altitude; but
if we can ignore this variation over the vertical dimension of the body, then the
body’s center of gravity is identical to its center of mass (abbreviated “cm”),
which we defined in Section 8.5. We stated this result without proof in Section 10.2,
and now we’ll prove it.

First let’s review the definition of the center of mass. For a collection of par-
ticles with masses m1, m2, c and coordinates 1x1, y1, z12, 1x2, y2, z22, c,

Also, xcm, ycm, and zcm are the components of the position vector r

Scm of the center

of mass, so Eqs. (11.3) are equivalent to the vector equation

Now consider the gravitational torque on a body of arbitrary shape (Fig. 11.2).
We assume that the acceleration due to gravity g
S
is the same at every point in
the body. Every particle in the body experiences a gravitational force, and the
total weight of the body is the vector sum of a large number of parallel forces. A
typical particle has mass mi and weight w
Si = mi gS. If rSi is the position vector of
this particle with respect to an arbitrary origin O, then the torque vector TSi of the weight wSi with respect to O is, from Eq. (10.3),

The total torque due to the gravitational forces on all the particles is

When we multiply and divide this by the total mass of the body,

The fraction in this equation is just the position vector rScm of the center of mass, with components xcm , ycm , and zcm , as given by Eq. (11.4), and MgS is equal to the total weight wS of the body. Thus

The total gravitational torque, given by Eq. (11.5), is the same as though the total weight wS were acting at the position rScm of the center of mass, which we also call the center of gravity. If gS has the same value at all points on a body, its center of gravity is identical to its center of mass. Note, however, that the center of mass is defined independently of any gravitational effect.

While the value of gS varies somewhat with elevation, the variation is ex-tremely slight (Fig. 11.3). We’ll assume throughout this chapter that the center of gravity and center of mass are identical unless explicitly stated otherwise.

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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: A body that can be modeled as a particle is in equilibrium whenever the vector sum of the forces acting on it is zero. But for the situations we’ve just described, that condition isn’t enough. If forces act at different points on an extended body, an additional requirement must be satisfied to ensure that the body has no tendency to rotate: The sum of the torques about any point must be zero. This requirement is based on the principles of rotational dynamics view more..
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Ans: Some physical quantities, such as time, temperature, mass, and density, can be described completely by a single number with a unit. But many other important quantities in physics have a direction associated with them and cannot be described by a single number. 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: 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: 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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