Statics and Equilibrium - How it works

Equilibrium and Vectors

Essential to calculations in statics is the use of vectors, or quantities that have both magnitude and direction. By contrast, a scalar has only magnitude. If one says that a certain piece of property has an area of one acre, there is no directional component. Nor is there a directional component involved in the act of moving the distance of 1 mi (1.6 km), since no statement has been made as to the direction of that mile. On the other hand, if someone or something experiences a displacement, or change in position, of 1 mi to the northeast, then what was a scalar description has been placed in the language of vectors.

Not only are mass and speed (as opposed to velocity) considered scalars; so too is time. This might seem odd at first glance, but—on Earth at least, and outside any special circumstances posed by quantum mechanics—time can only move forward. Hence, direction is not a factor. By contrast, force, equal to mass multiplied by acceleration, is a vector. So too is weight, a specific type of force equal to mass multiplied by the acceleration due to gravity (32 ft or [9.8 m] / sec 2 ). Force may be in any direction, but the direction of weight is always downward along a vertical plane.


Adding scalars is simple, since it involves mere arithmetic. The addition of vectors is more challenging, and usually requires drawing a diagram, for instance, if trying to obtain a vector sum for the velocity of a car that has maintained a uniform speed, but has changed direction several times.

One would begin by representing each vector as an arrow on a graph, with the tail of each vector at the head of the previous one. It would then be possible to draw a vector from the tail of the first to the head of the last. This is the sum of the vectors, known as a resultant, which measures the net change.

Suppose, for instance, that a car travels north 5 mi (8 km), east 2 mi (3.2 km), north 3 mi (4.8 km), east 3 mi, and finally south 3 mi. One must calculate its net displacement—in other words, not the sum of all the miles it has traveled, but the distance and direction between its starting point and its end point. First, one draws the vectors on a piece of graph paper, using a logical system that treats the y axis as the north-south plane, and the x axis as the east-west plane. Each vector should be in the form of an arrow pointing in the appropriate direction.

Having drawn all the vectors, the only remaining one is between the point where the car's journey ends and the starting point—that is, the resultant. The number of sides to the resulting shape is always one more than the number of vectors being added; the final side is the resultant.

In this particular case, the answer is fairly easy. Because the car traveled north 5 mi and ultimately moved east by 5 mi, returning to a position of 5 mi north, the segment from the resultant forms the hypotenuse of an equilateral (that is, all sides equal) right triangle. By applying the Pythagorean theorem, which states that the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides, one quickly arrives at a figure of 7.07 m (11.4 km) in a northeasterly direction. This is the car's net displacement.

Calculating Force and Tension in Equilibrium

Using vector sums, it is possible to make a number of calculations for objects in equilibrium, but these calculations are somewhat more challenging than those in the car illustration. One form of equilibrium calculation involves finding tension, or the force exerted by a supporting object on an object in equilibrium—a force that is always equal to the amount of weight supported. (Another way of saying this is that if the tension on the supporting object is equal to the weight it supports, then the supported object is in equilibrium.)

In calculations for tension, it is best to treat the supporting object—whether it be a rope, picture hook, horizontal strut or some other item—as though it were weightless. One should begin by drawing a free-body diagram, a sketch showing all the forces acting on the supported object. It is not necessary to show any forces (other than weight) that the object itself exerts, since those do not contribute to its equilibrium.


As with the distance vector graph discussed above, next one must equate these forces to the x and y axes. The distance graph example involved only segments already parallel to x and y, but suppose—using the numbers already discussed—the graph had called for the car to move in a perfect 45°-angle to the northeast along a distance of 7.07 mi. It would then have been easy to resolve this distance into an x component (5 mi east) and a y component (5 mi north)—which are equal to the other two sides of the equilateral triangle.

This resolution of x and y components is more challenging for calculations involving equilibrium, but once one understands the principle involved, it is easy to apply. For example, imagine a box suspended by two ropes, neither of which is at a 90°-angle to the box. Instead, each rope is at an acute angle, rather like two segments of a chain holding up a sign.

The x component will always be the product of tension (that is, weight) multiplied by the cosine of the angle. In a right triangle, one angle is always equal to 90°, and thus by definition, the other two angles are acute, or less than 90°. The angle of either rope is acute, and in fact, the rope itself may be considered the hypotenuse of an imaginary triangle. The base of the triangle is the x axis, and the angle between the base and the hypotenuse is the one under consideration.

Hence, we have the use of the cosine, which is the ratio between the adjacent leg (the base) of the triangle and the hypotenuse. Regardless of the size of the triangle, this figure is a constant for any particular angle. Likewise, to calculate the y component of the angle, one uses the sine, or the ratio between the opposite side and the hypotenuse. Keep in mind, once again, that the adjacent leg for the angle is by definition the same as the x axis, just as the opposite leg is the same as the y axis. The cosine (abbreviated cos), then, gives the x component of the angle, as the sine (abbreviated sin) does the y component.

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