### Video Transcript

A particle is accelerated when
acted on by two forces, as shown in the accompanying diagram. The force ๐น sub ๐ต has twice the
magnitude of force ๐น sub ๐ด. Find the direction in which the
particleโs net acceleration occurs in terms of the angle below the negative
๐ฅ-direction from the position of the particle.

So given the two forces acting on
the particle, ๐น sub ๐ต and ๐น sub ๐ด, we want to solve for a direction, the
direction in which the particleโs net acceleration occurs. Weโll call this direction ๐, where
๐ is an angle measured from the negative ๐ฅ-direction from the position of the
particle. In order to solve for the direction
of the particleโs acceleration, weโll want to solve for the resultant force acting
on the mass ๐. If we call that resultant force ๐น
sub ๐
, we know that force will have two components. One we can call ๐น sub ๐ฅ in the
๐-direction, that is, along the ๐ฅ-axis. And the second component we can
call ๐น sub ๐ฆ in the ๐-direction along the ๐ฆ-axis.

To solve for ๐น sub ๐
, the
resultant force, weโll add together ๐น sub ๐ต and ๐น sub ๐ด, the two forces acting
on the mass ๐. We can start by writing the force
vector ๐น sub ๐ด as the magnitude of that force multiplied by its direction in the
๐-direction. Likewise, we can write the force
vector ๐น sub ๐ต as negative the magnitude of that force multiplied by the cos of 45
degrees โ thatโs the ๐-component of ๐น sub ๐ต โ minus the magnitude ๐น sub ๐ต times
the sin of 45 degrees โ thatโs the ๐-component of that force.

In the exercise statement, weโre
told something about the magnitude ๐น sub ๐ต. Weโre told that itโs equal to twice
the magnitude of ๐น sub ๐ด. We substitute in two ๐น sub ๐ด for
๐น sub ๐ต and factor it out along with the minus sign from this expression for the
force vector ๐น sub ๐ต. We can further simplify this
expression for ๐น sub ๐ต by realizing that the cos of 45 degrees and the sin of 45
degrees are both the square root of two over two. And we see that now the factors of
two cancel out, so that ๐น sub ๐ต, the vector, is equal to negative root two the
magnitude of ๐น sub ๐ด in the ๐- and the ๐-direction.

Our next step is to add these two
forces, ๐น sub ๐ด and ๐น sub ๐ต, together. We find that their sum, which is
the resultant force ๐น sub ๐
, is equal to the magnitude of ๐น sub ๐ด times one
minus the square root of two ๐ minus the square root of two ๐. Weโve now solved for the resultant
or net force acting on our particle. And if we were to sketch in
approximately the direction of this force on our diagram, it might point in a
direction like this, slightly farther towards the negative ๐ฆ-direction and ๐น sub
๐ต. The particular direction this force
and therefore the particleโs acceleration is directed is given by the angle ๐
measured from the negative ๐ฅ-axis with respect to the particleโs position. If we were to draw in the ๐ฅ- and
๐ฆ-components of that resultant force, ๐น sub ๐
, on our diagram, we can see that
the angle ๐ is part of that right triangle. And we can see that the tangent of
that angle ๐ is equal to the ๐ฆ-component of the resultant force ๐น sub ๐ฆ divided
by the ๐ฅ-component ๐น sub ๐ฅ.

Looking at the resultant force we
solved for earlier, we see that the ๐ฆ-component of that force is the magnitude ๐น
sub ๐ด times negative root two. And we see further that the
๐ฅ-component of that force is the magnitude of ๐น sub ๐ด multiplied by one minus the
square root of two. Looking at this fraction, we see
that the magnitudes ๐น sub ๐ด cancel out. And if we then take the inverse
tangent of both sides of the equation, we find that ๐ is equal to the inverse
tangent of negative the square root of two divided by one minus the square root of
two. Entering this expression on our
calculator, we find that, to two significant figures, ๐ is 74 degrees. Thatโs the direction of the
particleโs acceleration as measured from the negative ๐ฅ-axis with respect to the
particleโs position.