OBJECT: To determine by means of vector diagrams the resultant of several concurrent forces and to check the accuracy of the result on a force table.

METHOD: Each student is assigned a problem. In this problem it is assumed that the magnitude and direction of certain forces acting on a body are known. The resultant and anti-resultant (equilibrant) of these given forces are determined both by graphical method and by the use of trigonometric relationships. The accuracy of the result is checked on a force table.

THEORY: Measurable quantities may be classified as either (I) scalar quantities or (2) vector quantities. A scalar quantity has magnitude only but a vector quantity has both magnitude and direction. For example, since to specify completely the velocity of a body it is necessary to state not only how fast it is traveling but in what direction it is going, velocity is a vector quantity. However, the mass of a body is completely specified by a magnitude and mass is a scalar quantity. Since the weight of a body is the force with which it is attracted by the earth, weight is a vector quantity.

Since weight and mass are different physical concepts, they should not be measured in the same units. The gram is a unit of mass. The force with which the earth attracts a one gram mass is called a gram weight of force.

Fig. 1 illustrates the method of representing vector quantities. Assume that the vector quantities here considered are forces, but the method applies equally well for all vector quantities. A symbol used to indicate the magnitude and direction of a vector quantity is called a vector. In this case the arrow A is a vector representing a force. The length of the line oa is drawn to scale to indicate the magnitude of the force. For example, 1cm might be taken to represent 200 gram weight of force. The direction of the line oa indicates the direction of the force and the arrow head indicates the sense, in this case from left to right. Similarly, the vectors B and C represent other forces.

If two or more forces are concurrent (meet at a common point) they may be replaced by a single force called the resultant.

One method of finding the resultant is indicated in Fig. 1. Having drawn the vectors A and B to scale and in the proper directions, the parallelogram oamb is constructed. The diagonal M of this parallelogram represents the resultant of the forces A and B. The scale previously chosen is used to determine the magnitude of the resultant and a protractor to determine the angle it makes with some chosen direction (the positive direction of A, for example). Since M may be used to replace A and B it should be obvious that M may be combined with C to find the resultant of A, B, and C. This method is known as the parallelogram method. The resultant of two or more forces is sometimes called the vector sum of the forces.

An alternative method of finding the resultant, called the polygon method, is illustrated in Fig. 2. The vector A is first constructed and from the head of A the vector B is drawn. Since the triangle formed by the vectors A, B, and M, in Fig. 2 and the triangle oam in Fig. 1 are identical, the vector M is the resultant of A and B. Similarly, the vector R is the resultant of A, B, and C. When the resultant of several forces is required, this method is somewhat simpler than the parallelogram method. It should be noted that when the parallelogram method is used, the arrows, with their tails together, all radiate from a common point, but that in the case of the polygon method the tail of the second arrow coincides with the head of the first, etc.

The resultant may also be determined analytically with the aid of trigonometric relationships. Applying these relations to Fig. 2 gives

M² = A² + B² + 2ABcosβ (1)
and
tanθ = Bsinβ/(A + Bcosβ) (2)

In many types of problems it is simpler to start by first resolving each force into components. In Fig. 3 the force B is the resultant of forces B_x and B_y and, therefore, conditions are not disturbed by replacing the single force B by these two forces. B_x and B_y are said to be the components of B. It is customary to determine the components of a force along mutually perpendicular axes, usually horizontal and vertical. It is obvious in Fig. 3 that B_x = Bcosβ and B_x = Bsinβ. Fig. 4 illustrates the component method of computing the resultant of A, B, and C. The X-axis is so chosen that it coincides with the vector A and the vectors B and C are resolved into X- and Y -components. The three forces A, B, and C have been replaced by five forces (A has no Y -component). The sum of the component along either axis may be computed by algebraic addition. Calling the sum of the X-components F_x and the sum of the Y-components F_y, it follows that the resultant R is given by the equation

R² = (F_x)² + (F_x)² (3)

and that the angle φ-the angle that R makes with the X-axis- may be determined from the equation

tanφ = F_y/F_x (4)

A body is said to be in equilibrium if it has no acceleration. A body that has no acceleration may be either at rest or moving with uniform speed in a straight line.

This experiment treats with a body at rest. From Newton’s second law it follows that if a body is in equilibrium the resultant of all the forces acting on it must be zero. If the forces acting on a body have a resultant other than zero, the body may be put into equilibrium by adding a force equal and opposite to the resultant force. This force is called the anti-resultant or equilibrant force.

APPARATUS:
The apparatus consists of a protractor, parallel ruler, and a force table. The force table is illustrated in Fig 5. The circular tabletop, calibrated in degrees, is mounted on a vertical rod held in a heavy tripod base equipped with leveling screws as shown in the illustration. The body whose equilibrium is under study is the ring at the center of the table. The central pin holds this ring in position when the weights are unbalanced. The forces acting on this ring are the tensions in the cords. If the friction in the pulleys is negligible the tensions in the cords are equal to the downward pull of gravity on the suspended masses. Each pulley clamp has an index by means of which the direction of the corresponding force may be read on the circular scale.

PROCEDURE: Each student is assigned one of the problems in Table 1 (or a similar problem) by the instructor

1. Using the parallelogram method, determine graphically the resultant and equilibrant of A and B. On the same diagram, determine the resultant and equilibrant of A, B, and C. Choose a scale such that the finished vector diagram will almost fill the sheet of paper, Use a sharp pencil and construct as accurately as the instruments will permit. Check both of the above results on the force table. In each case, when the adjustments on the force table have been completed, ask the instructor to check the results.
2. Using the polygon method determine graphically the resultant of A, B, and C. Compare this result with the one obtained above.
3. Using Eqs. (1) and (2) determine the resultant of B and C. Check the result on the force table.
4. Use the component method, illustrated in Fig. 4, and trigonometric relationships to compute the resultant of A, B, and C. Compare with the results above.
5. Determine experimentally on the force table (without the aid of vector diagrams) the equilibrant of A and C.

QUESTIONS:

1. The forces in this experiment act on a ring but are said to be concurrent. Explain. If the cords were attached rigidly to the ring would the forces necessarily be concurrent?
2. Indicate whether each of the following is a vector or scalar quantity: speed, velocity, mass, weight, work, torque, volume.
3. A hammock is supported by two hooks at the same level. A man is seated in this hammock. Under what conditions will the pull on each hook be equal to the man’s weight?
4. A body, weight W, is attached by a string, length l, to a hook on a vertical wall. A horizontal force F acting on the body holds it at a distance d from the wall. Derive the equation which gives the force F in terms of W, l, and d.
5. If two forces A and B are in line the resultant is A±B. Show how this follows from Eq. 1.

Equipment List

• Economy Force Table

From: Cenco Physics Selected Experiments in Physics (No. 71990-131), Copyright, 2003, Sargent-Welch Scientific Company.

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