Video Transcript
Use the following diagram to
find the tension in 𝐶𝐵. Round your answer to two
decimal places.
Remember, if a pair of forces
are acting on a rigid body and that body is in equilibrium, then a third force
𝐅 must also be acting on the body, which is equal in magnitude and opposite in
direction to the resultant of those two forces. In this case, the pair of
forces are the tensions with magnitudes 𝑇 sub one and 𝑇 sub two. The third force is the downward
force of 10 newtons, meaning that the magnitude of the resultant 𝑅 must also be
10 newtons.
So if we know the angle between
the two forces, let’s call that 𝜃, we can use this equation to find the
magnitude of the resultant. Substituting what we know about
this system and we get 10 equals the square root of 𝑇 sub one squared plus 𝑇
sub two squared plus two times 𝑇 sub one times 𝑇 sub two cos 𝜃.
Since in this case the triangle
is equilateral, 𝑇 sub one and 𝑇 sub two are equal. So we can replace these with
𝑇, square both sides, and begin to evaluate the right. We can now factor two 𝑇
squared, so 100 equals two 𝑇 squared times one plus cos 𝜃. Making 𝑇 squared the subject
by dividing through by two times one plus cos 𝜃 gives 𝑇 squared equals 50 over
one plus cos 𝜃.
Next, we might spot that we can
find the value of 𝜃 using the length of the sides in the triangle. We label our triangle as
shown. And we get 50 squared equals 30
squared plus 30 squared minus two times 30 times 30 cos 𝜃. That’s 2500 equals 1800 minus
1800 cos 𝜃, and we can rearrange to find that cos of 𝜃 equals negative seven
over 18. Let’s substitute that into our
earlier expression. That’s 𝑇 squared equals 50
over one plus negative seven over 18, which is 900 over 11.
Finally, we know that the
tension in 𝐶𝐵 is 𝑇 sub one, which is equal to 𝑇, so we just need to square
root. The square root of 900 over 11
is 9.0453. Correct to two decimal places,
that’s 9.05, so the tension in 𝐶𝐵 is 9.05 newtons.
