### Video Transcript

Three forces β π
one equals
negative five π’ minus two π£ plus two π€ newtons, π
two equals π£ minus three π€
newtons, and π
three equals negative π’ minus five π£ minus two π€ newtons, where
π’, π£, and π€ are three mutually perpendicular unit vectors β acted on a body for
three seconds. Find the magnitude of their
combined impulse on the body.

In this exercise, we have some
body, and weβre told itβs being acted on at the same time by three forces: π
one,
π
two, and π
three. These forces are all constant, and
weβre told that they act over a time interval weβll call Ξπ‘ of three seconds. We want to know the magnitude of
the impulse on the body from these three forces acting together.

We could calculate the impulse due
to each of the three forces individually and then add those impulses together. But a shorter way is to add our
three forces together as vectors to solve for the net force acting on our body. So we write out π
one, π
two, and
π
three, lining them up by their π’-, π£-, and π€-components. We then add them by their
components, starting with the π’-component. Negative five π’ minus π’ gives a
total of negative six π’. Then negative two π£ plus π£ minus
five π£ adds up to negative six π£. And lastly, two π€ minus three π€
minus two π€ equals negative three π€. The units of these components are
all newtons. And this vector overall is equal to
the net force acting on our body.

Knowing this, letβs recall a
mathematical relationship for the impulse experienced by an object due to a force
acting on it. Impulse π’ is equal to π
times
Ξπ‘. In other words, the impulse an
object experiences equals the net force acting on that object times the time
interval over which that force acts. In our case, we know π
sub net and
we also know Ξπ‘. That was given as three
seconds.

But letβs recall that we want to
solve for impulse magnitude. That means weβll need to calculate
the magnitude of our net force and use that in our equation. Clearing a bit of space, we can
recall that given a three-dimensional vector β weβll call this example vector π― β
the magnitude of π― is equal to the square root of its π’-component squared plus its
π£-component squared plus its π€-component squared.

Applying this relationship to our
net force, we see itβs equal to the square root of negative six squared plus
negative six squared plus negative three squared newtons. This is equal to the square root of
36 plus 36 plus nine newtons or the square root of 81 newtons, which is nine
newtons. Thatβs the magnitude of the net
force acting on our body. To calculate πΌ then, we substitute
this value in for the magnitude of π
sub net and Ξπ‘ is equal to three seconds. So when we calculate πΌ, we find a
result of 27 newton seconds. Thatβs the magnitude of the total
impulse acting on our body.