dition of no translational acceleration the up forces must be equal to the down forces, or P+Q=R. While to satisfy the second condition, that the moments of the forces shall balance, we have Pr=Qy, for these are the moments about the point B, and R has zero moment about that point. Or we may take moments about A and find Rx=Q(x+y). If moments are taken about any point other than A, B, or C, there will be three moments to reckon. If, for example, the point D is taken as the axis, the clockwise moment of Q must be equal to the sum of the counter-clockwise moments of P and R.

A

Χ

D B

FIG. 23.

с

61. Parallel Forces in General.-Any case of equilibrium with parallel forces may be discussed in a similar way, two conditions being met, namely, the sum of the forces acting in any one direction must be equal to the sum of the forces in the opposite direction, and the sum of the clockwise moments about any axis must be equal to the sum of the counter-clockwise moments.

62. Illustration.-A certain bar having no weight is acted on by four forces as shown in figure 24, forces of 4 lbs. and 2 lbs. acting upwards and 3 lbs. and 5 lbs. acting downwards, and it is required to find the single force necessary to produce equilibrium and the point on the bar where it must be applied. Since the total upward force is 6 while the downward force is 8, the required force F must be an upward force 2 to satisfy the first condition of no translational acceleration.

F=2

2

C

1 ft.

1 ft.

1 ft.

3

5

FIG 24.

This force must be applied at such a point on the bar as to satisfy the second condition, and make the clockwise moments balance the counter-clockwise moments about any axis. Take an axis through P, for instance. The moments about P are

therefore to produce equilibrium the applied force 2 must produce a clockwise moment 4. Since it must also act upward, it must be applied

at a distance 2 to the left of P, and consequently the bar must be extended two feet in that direction.

Any point whatever on the bar might have been taken as the origin of moments, and the reader should show that the same conclusion is reached taking moments about some point such as C.

63. Couple and Torque. If in the case just treated the upward force 4 is changed to 6, we have a case that calls for special consideration. The upward forces. are exactly equal to the downward. forces and yet the bar is not in equilibrium, for taking

moments

3 1 ft.

1 ft.

1 ft.

5

FIG. 25.

6

about P we find that the clockwise moment is 10 while the counter-clockwise moment is 2+18=20. Here, then, there is a combination of forces that does not tend to produce translation, but simply rotation. Such a combination is known. as a couple, and its moment is commonly known as a torque. It cannot be balanced by any single force, for any force applied either upward or downward would cause translation. A couple can be balanced only by another couple having an equal and opposite moment, or torque.

The simplest case of a couple is when two equal parallel forces act in opposite directions not in the same straight line. For instance, the forces F F, figure 26, constitute a clockwise couple the moment of which is Fx where x is the distance between the lines of action of the forces. The moment of a couple about any axis is the same as about any parallel axis. For, take an axis perpen dicular to the paper and through P at a distance y y from the nearer force, then the moments are Fy P counter-clockwise and F(x+y) clockwise, hence subtracting we have Fx clockwise, as the resultant of the two.

F

FIG. 26.-Couple.

The moment of such a couple about an axis perpendicular to the plane in which the two forces lie is therefore measured by the product of the amount of either force by the perpendicular distance between their lines of action.

To produce equilibrium, then, in the case under consideration a couple having a clockwise moment 10 must be applied to the

bar, and it may be applied at any point we choose. The following figure illustrates different modes of producing equilibrium in this case.

In every case of equilibrium the forces acting may be resolved into a number of balancing couples.

5

3

2 ft.

10 1 ft.

6

5

10

21

6

5

5

64. Center of Gravity.-The weight of a mass is the force with which it is drawn toward the earth. All parts of a body have weight and so the weights of the several parts into which a body may be conceived to be divided constitute a system of parallel forces acting downward toward the earth. The resultant of this system of forces is a single force equal to their sum and is the

FIG. 27.

total weight of the body. It may be proved that there is a certain point in the body through which the resultant force due -- to weight always acts whatever may be the position of the body.

This point is called the center of gravity of the body. In all problems that involve the weight of a body we may ignore the fact that the weight is distributed throughout the body, and treat it as a single force applied at the center of gravity.

m

b

x P M a

B

FIG. 28.

65. Proof of Center of Gravity.-Let M and m be the masses of two parts of a body and let the line joining them be inclined as shown in figure 28. Since the weights of masses are proportional to the masses themselves ($38), the single upward force necessary to balance the weights of the two masses must be applied in the vertical line AB, so situated that Mx But AB intersects at P the line joining the two masses, dividing it into the two segments a and b which, by similar triangles, are in the same ratio as x and y, and consequently a: b:: M: m; and since this ratio does not depend on the inclination of the line joining M and m, it

=

my.

follows that the balancing force must pass through the point P whatever the inclination may be. P is therefore the center of gravity of M and m. Now conceive the masses M and m concentrated at P and find similarly a point P' through which the resultant weight of (M+m) and of another mass m' must always pass. Continue in this manner until account has been taken of all the masses into which we may conceive the body to have been subdivided. The point through which the final resultant passes is the center of gravity of the body.

66. Center of Mass and of Inertia. The center of gravity as has just been explained is determined by the distribution of mass in a body or system of bodies. It has certain remarkable properties quite independent of weight, and is therefore also called the center of mass or center of inertia of the body or system.

For example, a freely rotating body like a spinning projectile will always rotate about an axis through it center of mass.

FIG. 29.

67. Position of Center of Gravity. When a body is hung by a cord or balanced on a point the center of gravity must be in the vertical line passing through the point of support. For two equilibrating forces must act in the same straight line. If, therefore, a body is hung first from one point and then from another the intersection of the two lines thus determined marks the position of the center of gravity, as shown in the figure in case of a chair.

The center of gravity of a uniform bar is at its center; in a uniform thin plate, square, rectangular, or in the form of a parallelogram, it is at the intersection of the diagonals. In case of any homogeneous symmetrical body it lies in the plane of symmetry. Thus it is at the center of a sphere, circle, or circular disc and at the center of a cube.

68. Equilibrium under Gravity.-That a body may be in

FIG. 30.

equilibrium under gravity, it must be supported by a force equal to its own weight and acting upward through its center of gravity. Thus the two cones and sphere shown in the figure as resting on a level table are in equilibrium, the upward force being

But the first cone is said

supplied by the reaction of the table. to be in stable equilibrium, because if slightly tipped it will fall back to its original position. The second cone is said to be in unstable equilibrium because if disturbed it will fall away from its original position, while the sphere is said to be in neutral equilibrium because it remains in equilibrium when displaced. It will be observed that in the first case the center of gravity of the cone is raised when it is tipped; in the second it is lowered, and in case of the sphere it is neither raised nor lowered. The weight of a body being considered as acting at its center of gravity will always cause that point. to move down or toward the earth unless apposed by some other force. In case of a loaded wagon

B

FIG. 31.

on a side hill the vertical line through its center of gravity may remain between the wheels if the center of gravity is low, when if it were high the line of action of the weight might fall outside the wheel base causing the load to overturn.

69. Balances.-The beam of a balance rests on a sharp steel "knife edge" A, while the pans are hung on the knife edges B and C. These three knife edges are rigidly fixed in the beam and should be