The Equilibrium of Forces: The Second Law©98

Experiment 8

Objective:To study the methods of vector addition as applied to be equilibrium of torques.

Discussion:

The second law of the equilibrium of forces states that the vector sum of the torques, acting on a rigid unaccelerated object, is zero. The torque due to a force depends upon three things:

  1. The force acting on the body.
  2. The point at which this force acts.
  3. The point relative to which the torque is measured.

Since this latter point is arbitrary, that is, since it is not determined by the physical situation, the value of the torque is arbitrary. Nevertheless, the vector sum of the torques acting on a body in equilibrium is zero, provided all the torques have been measured relative to the same reference point.

The magnitude of the torque  due to a force is defined to be the product of the magnitude of the force F, the magnitude of the displacement r of the point of application of the force from the reference point, and the sine of the angle  between the vectors F and r. That is,

(1)

The direction of the torque is always perpendicular to the plane formed by the vectors r and F. For example, if r and F lie in the plane of this page, the torque is out of or into the page. All the forces and displacements we shall be considering in this laboratory lie in a common plane, their torques are either parallel or antiparallel to one another, lying perpendicular to the plane. In such a situation, it is more convenient to specify whether a torque is clockwise or counterclockwise. These ideas are diagramed in Fig. 1. This figure shows a body being acted upon by a force F. The point relative to which torques are measured is the point O, which is chosen arbitrarily, and for simplicity it was chosen to be the center of mass of the object. The torque is clockwise if the angle, when measured fromr toF is designated by an arrow pointing clockwise. The torque is counterclockwise if the arrow points in the counterclockwise direction. The torque inFig. 1 is counterclockwise.

Returning now to the second law of the equilibrium of forces, we translate its "vector" language into one using "clockwise" and "counterclockwise": the sum of the clockwise torques acting upon a rigid body must equal the sum of the counterclockwise torques if body is to be in equilibrium. Written thusly, this law makes sense, of course, only when all the forces, Fn, and the displacements, rn, lie in the same plane.

The apparatus in this experiment consists of a meter stick supported by two strings. Weights hang from these strings as well as from the meter stick. A diagram of the apparatus is shown in Fig. 2. The meter stick is the body on which the forces act, and the forces themselves are due to the weights hanging from the strings and meter stick. Since the forces do not all act at the same point on the meter stick, the torque due to these forces must balance if equilibrium is to be a maintained. In addition, the forces themselves must balance.

The supporting strings are not vertical, and so they pullhorizontally as well as vertically on the stick. In establishing whether or not the system is theoretically in equilibrium, we need to decompose the forces due to the supporting strings into their horizontal and vertical components. This decomposition is effected by constructing a righttriangle whose hypotenuse lies along the string and whose arms are horizontal and vertical, as seen in Fig. 3. LettingF be the force on the string, Fh its horizontal component andFv its vertical component, we have

(2)

Having decomposed the forces due to the supporting strings, we may test the first law of equilibrium, that the vector sum of the forces must be zero. Using the forces asshown in Fig. 2, we have

(3)

and

(4)

whereW is the weight of the meter stick acting at the center of the meter stick.

To test the second law of equilibrium, we measurer from somearbitrarily chosen point. In this example that point is zero-point of the meter stick. Hence all the r measurements were from the end of the meterstick in Fig. 2. Thus the torque equation is:

(5)

Here, the counterclockwise torques are positive and the clockwise torques are negative. Notice that there is no reference to sin1, sin3 or sin4; all these angles are 90° and the sine of 90° is 1. Since the point from which the torques are measured is immaterial, we could just well have measured them from say, the middle of the meter stick as long as we are careful about the signs.

Exercises:

  1. Determine the forces on the meter stick.
  2. Weigh the meter stick and determine the position of its center of gravity.
  3. Add weights to the supporting strings and meter stick so that the meter stick is in equilibrium when it is horizontal.
  4. Measure the forcesF1,F2,F4andF5, by noting the masses of the weights hanging from the supporting string and meter stick.
  1. Determine the h- and v-components of the forces.
  2. Measure the angle of elevation each string makes with the horizontal.
  3. On paper, construct right triangles whose base angles equal the angles of elevation of the strings.
  4. Letting the hypotenuses of the right triangles be proportional to the tensions in the strings, measure the arms of the triangles to find the horizontal and vertical components of the tension.
  1. Find the percent deviation of the sum of the upward forces from the sum of the downward forces by inserting the measured values of the vertical components of the forces into Eq. (3).
  1. Find the percent deviation of the sum of the forces in the +x-direction from the sum of the forcesin the -x-direction by inserting the measured values of the horizontal components into Eq. (4).
  1. Analyze the torques.
  2. Calculate the torques exerted by each of the forces(measuringr from the zero-point of the meter stick).
  3. Label whether the torques are clockwise or counterclockwise.
  4. Find the percent deviation of the sum of the clockwise torques from the sum of the counterclockwise torques using Eq. (5).
  1. Repeat Exercise 5, measuringr from the center of gravity of the meter stick.

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