Video Transcript
Two perpendicular forces, 𝐅 sub
one and 𝐅 sub two, act at a point. Their resultant, 𝐑, has magnitude
188 newtons and makes an angle of 60 degrees with 𝐅 sub one. Find the magnitudes of 𝐅 sub one
and 𝐅 sub two.
We begin by drawing a sketch of the
two perpendicular forces 𝐅 sub one and 𝐅 sub two that act at a point that we will
call 𝑃. We are told that the resultant of
these two forces has a magnitude of 188 newtons and that this resultant force makes
an angle of 60 degrees with 𝐅 sub one. By creating a right triangle as
shown, we can use the trigonometric ratios and the Pythagorean theorem to calculate
the magnitudes of 𝐅 sub one and 𝐅 sub two.
The Pythagorean theorem states that
𝑎 squared plus 𝑏 squared equals 𝑐 squared, where 𝑐 is the length of the
hypotenuse and 𝑎 and 𝑏 are lengths of the two shorter sides of any right
triangle. In this question, this means that
the magnitude of 𝐅 sub one squared plus the magnitude of 𝐅 sub two squared is
equal to the magnitude of 𝐑 squared. And as such, the magnitude of 𝐅
sub one squared plus the magnitude of 𝐅 sub two squared is equal to 188 squared,
which is equal to 35,344. We will call this equation one.
Next, we recall that in any right
triangle the tangent of any angle 𝜃 is equal to the opposite over the adjacent. From our diagram, this means that
tan of 60 degrees is equal to the magnitude of 𝐅 sub two over the magnitude of 𝐅
sub one. From our knowledge of the special
angles, we know that tan of 60 degrees equals root three. So rearranging our equation, we
have root three multiplied by the magnitude of 𝐅 sub one is equal to the magnitude
of 𝐅 sub two. We can then square both sides of
this equation, giving us three multiplied by the magnitude of 𝐅 sub one squared is
equal to the magnitude of 𝐅 sub two squared. We will call this equation two. And we now have a pair of
simultaneous equations with two unknowns, 𝐅 sub one and 𝐅 sub two.
We can now substitute equation two
into equation one such that 𝐅 sub one squared plus three 𝐅 sub one squared is
equal to 35,344. The left-hand side simplifies to
four 𝐅 sub one squared. We can then divide through by four
and square root both sides of the equation such that the magnitude of 𝐅 sub one is
equal to 94 newtons. We can then substitute this value
back into equation two so that the magnitude of 𝐅 sub two squared is equal to three
multiplied by 94 squared. Three multiplied by 94 squared is
equal to 26,508. We can then square root both sides
of this equation so that the magnitude of 𝐅 sub two is equal to 94 root three
newtons.
We now have the two required
magnitudes of 𝐅 sub one and 𝐅 sub two. They are equal to 94 newtons and 94
root three newtons, respectively.
