A tractor has an engine of 187
horsepower and it is pulling against a force of 374 kilograms-weight. Find its maximum speed.
All right, so let’s say that this
is our tractor. And we’re told that it has an
engine with a power of 187 horsepower. It exerts this power to pull
against a force of 374 kilograms-weight. This is a way of telling us that
the force opposing the tractor’s forward motion is 374 kilograms-weight. For the tractor to move forward as
designed, it needs to match this force in the opposite direction. Its engine supplies the power to do
that. The tractor will move at a constant
speed when the forward force 𝐹 matches 374 kilograms-weight in magnitude.
And when all 187 horsepower of its
engine are used, it will be moving at its maximum speed. These three quantities — power,
force, and speed — are related by this equation. Since we want to solve for speed
rather than power, we can divide both sides by the force 𝐹, cancelling that out on
the right. This leaves us with the result that
𝑃 over 𝐹 equals 𝑣. In theory, we could solve for the
maximum speed of the tractor by inputing our power and our force to this
The trouble, though, is we would
end up with units of horsepower per kilograms weight, an unfamiliar unit for
speed. To express the maximum speed in
more familiar units, we can convert horsepower and kilograms-weight into their
respective SI base unit equivalents. One metric horsepower is equal to
735 watts. And one kilogram-weight equals 9.8
newtons. So if we multiply the numerator of
our fraction by 735 watts per horsepower, then in terms of the units, horsepower
will cancel out and we’ll be left with units of watts. And if we likewise multiply our
denominator by 9.8 newtons per kilogram-weight, then kilogram-weight will cancel
out, leaving us with units of newtons.
Note that in each case, as we
multiplied by these conversion factors, we were effectively multiplying by one. That’s what makes these operations
allowable algebraically. When all the dust settles then, we
end up with units of watts per newton. And as we can infer from our
equation, power divided by force is equal to speed, a watt per newton is equal to a
meter per second.
We’re almost ready to calculate our
maximum speed 𝑣. Before we do, there’s one last
change to make. We could report 𝑣 in meters per
second, but given the units we started with, horsepower and kilograms-weight, it’s
more common to report speeds in kilometers per hour. Remembering that one meter per
second is equal to 3.6 kilometers per hour, if we multiply our fraction by 3.6
kilometers per hour per meter per second, effectively multiplying by one, then we
see the units of meters per second cancel from numerator and denominator.
Finally, then, we have an
expression for 𝑣 in the units of interest. Computing the speed, we find a
result of 135 kilometers per hour. This is the maximum speed the
tractor can attain.