Work and the Conservation of Energy 173
Look at this diagram. The magnitude of the force needed to push the load
up this ramp (F) is equal to the component of the force of gravity parallel
to the ramp (PR). So, if the ramp has a length of d, the work required to
move the load to height h can be represented as:
work = Fd
Now, you know intuitively that F is smaller than mg, and d is larger than h.
That makes sense. Is that why it takes the same amount of work to push
the load up a ramp as it does to lift the load straight up?
Yes, indeed. Now let’s show why this works, mathematically. r ABC repre-
sents the ramp in the figure, and rPQR represents the composition of the
force mg. These two triangles are similar—this means that ∠CAB = ∠RPQ.
This also means that the proportion of their corresponding sides must be
the same, as well. Thus, the following must be true:
Let’s make this a little less abstract. The line segment AB equals d (length
of ramp) and AC equals h (height). Similarly, the line segment PQ equals
mg (the force downward, due to gravity), while PR equals F (the force
applied to offset a portion of that force).