Question Video: Finding the Missing Term in the Vector Forces Expression Mathematics

Complete the following: In the given figure, 𝐅₁ = (𝐑 Γ— …)/(sin (πœƒβ‚ + πœƒβ‚‚)).

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Video Transcript

Complete the following. In the given figure, 𝐅 sub one is equal to 𝐑 multiplied by what divided by the sin of πœƒ sub one plus πœƒ sub two.

The diagram shows a parallelogram of forces, where the sides of the parallelogram are equal to the magnitudes of vectors 𝐅 sub one and 𝐅 sub two and the diagonal of the parallelogram is the resultant of these forces 𝐑. In order to find an expression for 𝐅 sub one, we can begin by finding the missing angles in triangle 𝑂𝐴𝐢. Angles 𝐡𝑂𝐢 and 𝑂𝐢𝐴 are alternate angles, meaning that angle 𝑂𝐢𝐴 is equal to πœƒ sub two.

We know that angles in a triangle sum to 180 degrees. This means that the measure of angle 𝑂𝐴𝐢 is equal to 180 degrees minus πœƒ sub one plus πœƒ sub two. We can now use the sine rule or law of sines to find an expression for 𝐅 sub one in terms of 𝐑 together with πœƒ sub one and πœƒ sub two. The law of sines states that π‘Ž over sin 𝐴 is equal to 𝑏 over sin 𝐡, which is equal to 𝑐 over sin 𝐢, where lowercase π‘Ž, 𝑏, and 𝑐 are the lengths of the three sides of our triangle and capital 𝐴, 𝐡, and 𝐢 are the measures of the angles opposite the corresponding sides.

In triangle 𝑂𝐴𝐢, side length 𝑂𝐴 is opposite the angle at 𝐢. This gives us 𝐅 sub one over sin of πœƒ sub two. The resultant 𝐑 is opposite the angle equal to 180 degrees minus πœƒ sub one plus πœƒ sub two. Substituting this into the law of sines gives us 𝐑 over sin of 180 degrees minus πœƒ sub one plus πœƒ sub two. Recalling the symmetry of the sine function, we know that the sin of 180 degrees minus 𝛼 is equal to sin 𝛼. This means that the right-hand side of our equation can be rewritten as 𝐑 over sin of πœƒ sub one plus πœƒ sub two. In order to make 𝐅 sub one the subject of our equation, we can multiply through by sin of πœƒ sub two. 𝐅 sub one is therefore equal to 𝐑 multiplied by sin of πœƒ sub two divided by sin of πœƒ sub one plus πœƒ sub two.

And we can therefore conclude that the missing part of the equation in the question is sin of πœƒ sub two.

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