# Sine 30 or Sine 60

I am looking for a second pair of eyes for verification. In the solutions manual provided by the textbook and professor, sin30 is used to calculate the answer. However, some of the students are arguing with the professor and everyone believes their answer is correct. I believe the textbook may be incorrect however, I decided to turn it over to the community for confirmation. This is a basic question to introduce the students to vectors mathematics. At this point in time, it appears seems that everyone's answer will be considered valid at this point. So I am asking the community who is unbiased what they think should be used to calculate the answer. I thank you for your time it is truly appreciated.

• My usual intuition in problems like this is that when computing a torque about an axis, the moment arm is determined by the distance of the line of action from the axis. (The "line of action" is the line through the point where the force is applied, in the direction of the force.) It does not matter where on the line the force occurs, so the force could just as well be applied at the point of closest approach to the axis as at any other point on the line. In this case, if you extend the force vector in a line it intersects the perpendicular radius at a distance $r\cos30^\circ$ from the axis. Sep 5, 2020 at 18:38

The solution is wrong. The definition of the magnitude of the cross product is $$|\vec a\times\vec b|=ab\sin\alpha,$$ where $$\alpha$$ is the angle between the vectors. But in your case the angle between the vectors is not $$30^\circ$$ but $$120^\circ$$, so the answer should be $$rF\sin 120^\circ=rF\cos\theta$$
• This formulation suggests a possible cause for the error in the book. The author recalled that the formula involves a factor $\sin \alpha,$ and having labeled just exactly one angle in the entire diagram, the author absent-mindedly chose that angle's value for $\alpha.$ Sep 5, 2020 at 18:27