Question Video: Resolving Forces | Nagwa Question Video: Resolving Forces | Nagwa

Question Video: Resolving Forces Mathematics • Second Year of Secondary School

Join Nagwa Classes

Attend live General Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

True or False: In the given figure, 𝐅₁/cos (90Β° βˆ’ ΞΈ) = 𝐅₂/sin ΞΈ = 𝐑/sin 90Β°

03:55

Video Transcript

True or False: In the given figure, 𝐅 sub one over the cos of 90 degrees minus πœƒ is equal to 𝐅 sub two over sin πœƒ, which is equal to 𝐑 over sin of 90 degrees.

The figure shows two perpendicular vector forces 𝐅 sub one and 𝐅 sub two, together with their resultant 𝐑 which acts at an angle of πœƒ degrees to the horizontal. Since the line segments 𝑂𝐡 and 𝐴𝐢 are parallel and equal in length and as vectors describe direction and magnitude but not location, we can let the line segment 𝐴𝐢 be equal to the vector 𝐅 sub two. We can now consider triangle 𝑂𝐴𝐢 in order to find the link between vector 𝐅 sub one, 𝐅 sub two, and their resultant.

Since 𝐅 sub one and 𝐅 sub two are perpendicular, we know that angle 𝑂𝐴𝐢 is a right angle and is therefore equal to 90 degrees. We know that angles in a triangle sum to 180 degrees. This means that the measure of angle 𝑂𝐢𝐴 is equal to 180 degrees minus 90 degrees plus πœƒ, which is equal to 90 degrees minus πœƒ.

Next, we can use the sine rule or law of sines. This 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 angles opposite the corresponding sides. In our triangle, vector 𝐅 sub one is opposite the angle 90 degrees minus πœƒ, 𝐅 sub two is opposite angle πœƒ, and the resultant force 𝐑 is opposite the angle of 90 degrees. Substituting in these values gives us 𝐅 sub one over sin of 90 degrees minus πœƒ is equal to 𝐅 sub two over sin πœƒ, which is equal to 𝐑 over sin of 90 degrees.

The second and third parts of our equation are correct. However, the denominator of the first part of the equation in the question was the cos of 90 degrees minus πœƒ, whereas in the figure the correct expression is the sin of 90 degrees minus πœƒ. From our knowledge of supplementary angles and the trigonometric identities, we know that the cos of 90 degrees minus πœƒ is equal to the sin of πœƒ, and this will only be the same as sin of 90 degrees minus πœƒ when πœƒ is 45 degrees.

If πœƒ was equal to 45 degrees, then 𝑂𝐴𝐢𝐡 would be a square, and the magnitude of 𝐅 sub one would be equal to the magnitude of 𝐅 sub two. As we are not told that this is the case in this question, we can conclude that the statement is false.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy