Forces at other angles

Work is only done by the component of the force parallel to the motion.

Suppose the force is at some other angle with respect to the motion, say θ = 45 °. Such a force could be broken down into two components, one along the direction of the motion and the other perpendicular to it. The force vector equals the vector sum of its two components, and the principle of vector addition of forces thus tells us that the work done by the total force cannot be any different than the sum of the works that would be done by the two forces by themselves. Since the component perpendicular to the motion does no work, the work done by the force must be

where the vector d is simply a less cumbersome version of the notation Δr. This result can be rewritten via trigonometry as

Even though this equation has vectors in it, it depends only on their magnitudes, and the magnitude of a vector is a scalar. Work is therefore still a scalar quantity, which only makes sense if it is defined as the transfer of energy. Ten gallons of gasoline have the ability to do a certain amount of mechanical work, and when you pull in to a full-service gas station you don't have to say Fill 'er up with 10 gallons of south-going gas.

Students often wonder why this equation involves a cosine rather than a sine, or ask if it would ever be a sine. In vector addition, the treatment of sines and cosines seemed more equal and democratic, so why is the cosine so special now? The answer is that if we are going to describe, say, a velocity vector, we must give both the component parallel to the x axis and the component perpendicular to the x axis (i.e., the y component). In calculating work, however, the force component perpendicular to the motion is irrelevant - it changes the direction of motion without increasing or decreasing the energy of the object on which it acts. In this context, it is only the parallel force component that matters, so only the cosine occurs.

Self-Check

(a) Work is the transfer of energy. According to this definition, is the horse in the picture doing work on the pack? (b) If you calculate work by the method described in this section, is the horse doing work on the pack? Answer, p. 158 Breaking Trail, by Walter E. Bohl.

Pushing a broom.

A violin.