General intuition and tips for solving vector problems in mechanics?

[Inertigratus]

Homework Statement

I'm just looking for ideas, like how to proceed with various problems and how to counter them.
Almost all problems I have encountered so far haven't been very much like the examples, but instead kind of vague and dependant on trigonometry and vectors.

Right now I'm stuck trying to solve problems about Moment of Force.

The thing is that I don't have a PhD in trigonometry so the questions are kind of tricky for me. They even have the answer (just the answer) below the question, just so they can say that "Well, the answer is right in front of you!".

So when you have problems in which you don't have any numbers to go on, maybe an angle and a side or two. For example if you have a force vector between A and B, and you have the length from O to A and from O to B as well as the angle between those two lengths.
Together the two lengths and the vector AB form a triangle (not a right triangle) and you're supposed to find the moment about point O "as a function" of the angle.

Homework Equations

Mo = r x F = F*r*sin(a)

The Attempt at a Solution

I know that if the arm isn't perpendicular to the force then you have to find the shortest distance between the point about which I'm trying to find the "torque" and that is given by the length of the arm times sine of the angle between the arm and the force's line of action.

 

[tiny-tim]

Hi Inertigratus!

I know that if the arm isn't perpendicular to the force then you have to find the shortest distance between the point about which I'm trying to find the "torque" and that is given by the length of the arm times sine of the angle between the arm and the force's line of action.

That's right!

So you know the lengths a and b (and maybe c?), and you also know the angle between them, and you have to find the sine of one of the other angles.

Using trig (there's also a Cartesian coordinate method, using the cross product ), you can find it by using the sine formula for triangles (sinA/a = sinC/c); if you don't know the third side of the triangle (c), you'll need to use the cosine formula also, to find that.