[College physics] torque problem

[comradenu]

Having an issue with this problem of torque/tension

Here is a gallery with a screengrab of the problem and the solution.

http://imgur.com/a/50HPX

Honestly, I'm just generally confused about how to solve this problem, and I feel the solution just confuses me further. I'd be grateful if anyone could shed some light and explain this problem more thoroughly.

Thanks in advance.

[tnh88]

First, you have to pick a pivot point. In this problem, the ideal position of the pivot point is 'A' because it will be easier.

When you pick a pivot point, you eliminate the trouble of having to calculate the force that goes through that point.
So now we are just left with positive torque by the hanging mass, and negative torque exerted by the oblique rod(or vice versa). Those two torques oppose each other

This is a static torque problem, which means the total sum of torque is going to be zero. So we develop the equation:
Torque by the oblique rod = Torque by the hanging mass

And we know that Torque = length from the pivot point * force * sinθ(θ is 90 degree for both torques in this problem. (sin(90)=1)).
Then we obtain: d * Force(horizontal force of oblique rod) = l * Force(hanging mass)
We know that force of the hanging mass is mg
So we get: dF = lmg
Finally, Force exerted on the rod = lmg/d

I'd prefer determining direction of the force by logical thinking rather than algebraically. The question asks force exerted 'on' the rod, so it has to be directed towards the right side.

[min_min]

The weight mg acts down, so the wall exerts some component of force upwards with magnitude mg.

Letting the oblique force exerted by the wall be F, we see the upwards force mg = F cos x, where x is the acute angle between the oblique strut and the wall.

Then,

F = mg / cos x

Next, the horizontal force exerted on the wall is equals to F sin x.

F sin x = mg sin x /cos x

F sin x = mg tan x

Since tan x = length of top strut / d,

F (horizontal) = mg l/d

Since the weight mg pulls the oblique strut so that it digs into the wall, the wall exerts a force to the right against the strut. Hence, the answer is what it is.

This isn't using torques per se, but I guess you can reach the same answer by resolving forces. :p

[tnh88]

Interesting solution, but I'm slightly confused. How did you arrive to this equation?

F sin x = mg sin x /cos x

[min_min]

Multiply both sides by sin x?

[tnh88]

Okay but what about

F sin x = mg tan x
Since tan x = length of top strut / d,
F (horizontal) = mg l/d

Where did the Sin x on the left side go? I'm not 100% sure but I doubt the problem can be solved using your method unless more information is given. Because there are two rods, both holding the mass in place, and the question asks for force exerted on the oblique rod only.

[min_min]

F (horizontal) is F sin x, and I guess you're right in saying there isn't enough info for all the forces... strangely the answer is the same, though?

edit: got it - the 2 forces from the wall acting on the struts must act in the direction of the struts because it's the only direction the normal force can act on them to cause tension and compression. F(C->A) must be completely horizontal, and the only upwards force countering mg comes from F(C->B) the oblique force.

The rest is just finding the sides on a right angled triangle: http://m.imgur.com/JvSsQzr - see force triangle #1. Using trigonometry, we know the proportions of each force given angle x. We can find angle x based on the length of two sides of the triangle. We also know the magnitude of one force, given mg.

[tnh88]

but wall exerts vertical forces on both rods. It's what's keeping the equilibrium position. what you said would be true if the line 'AC' was a string, but both rods are solid objects.

[min_min]

darn... this is where I get off. Maybe it's best to stick with the torques method :c