I'm gonna give this a shot. It's been a while since I looked at this stuff, so take it as a grain of salt:
Your first step should be to find a place to call the pivot point. Often the easiest choice would be where all the unknown forces are acting. Doing this would cancel them out.
Here are a list of forces in this problem:
- Fy,wall
- Fx,wall
- F,pole
- F,tension
- F,object
In this case, the three forces in red are the "unknowns" that you don't need (assuming you're told the mass of the object and asked to find the tension). This is a hinge problem, so most likely the pivot point will be point A. Choosing this as our pivot, we can eliminate the force of the wall (both x and y components). Also, by assuming the pole is massless we could also eliminate it's torque.
This is exactly where you want to be.
You're now left with TWO torques to consider:
- The Tension in the String (acting some distance from the pivot)
- The weight of the Object (acting some distance from the pivot)
Because these are the only two (non-zero) torques, and because everything is in static equilibrium, these two torques must be balancing (and therefore pointing in opposite directions).
We could now say the following:
- The Tension in the String (pulling counter-clockwise)
- The weight of the Object (pulling clockwise direction)
Net Torque = 0 (Net Torque = Torque of Tension - Torque of Object);
Ftension x (Lever Arm) = Fobject x (Lever Arm)
The main problem here is finding the distances to which the forces acting. To do this, you need to consider trigonometry.
This part is a little more difficult to explain. The theta position in the upper triangle (at point A) is equivalent to the theta of the bottom angle (at point C).
The distance from A to B is:
Cos(theta) = Adjacent / Hypotenuse
Cos(theta) x Hypothenuse = Adjacent
"Adjacent" being the length from A to B
The distance from A to (bottom of) C is:
Sin(theta) = Opposite / Hypotenuse
Sin(theta) x Hypotenuse = Opposite
"Opposite" being the Length from A to C
Here did you mean B to C? A to C is the hyp no?
Now plug this into the original torque equation we had earlier:
Ftension x (Lever Arm) = Fobject x (Lever Arm)
Ftension x (Cos(theta) x Hypothenuse) = Fobject x (Sin(theta) x Hypotenuse = Opposite)
Since you didn't specify what the weight of the object is, the tension, or even the angle, there's not much more you can do really.
But let's say you were trying to solve for tension:
You'd have to re-arrange the equation to look like this:
Ftension = Fobject x Sin(theta)/Cos(theta) <--- "Hypotenuse" cancels
Ftension = Fobject x tan(theta)
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Note: If the pole wasn't massless, you could still find the tension. The torque would balance out a bit differently.
Instead of: Ftension x (Lever Arm) = Fobject x (Lever Arm)
You'd have: Ftension x (Lever Arm) = Fpole x (Lever Arm) + Fobject x (Lever Arm)
Or simply put: Torque (Tension) = Torque (Pole) + Torque (Object)
Assuming the pole has uniform mass, the lever arm for the pole would be half the distance of the Lever Arm to the object (due to the pole's center of mass).
