Suspended Beam: Solving Homework with Unknown Equations

[PierceJ]

Homework Statement

Homework Equations

I do not know exactly.

The Attempt at a Solution

I honestly have no idea how to approach this.

 

[axmls]

Do you know what the forces acting upon the beam are?

 

[haruspex]

State a law relating forces to accelerations.

 

[PierceJ]

Do you know what the forces acting upon the beam are?

Gravity and the Tension in the wire

 

[axmls]

It can't be just those. Gravity points downwards, and the tension force points up and to the left. Since this beam is at rest, you've got to have something canceling that leftward force out.

Here's a hint. The hinge produces two reaction forces: a horizontal and vertical component. Take those into account, draw a free body diagram, and write the expressions for the net forces and moments.

 

[PierceJ]

Torque = Fd
Tsinθ(2.41) - mg(2.9) = 0
Find the angle through trig?

 

[axmls]

Look at the hint again and factor that into the equations. You will need to find the net force in the x-direction and the net force in the y-direction. You don't need the angle, as you are given the sides of the triangle formed by the cable.

 

[PierceJ]

I don't understand why I don't need the angle. Can I get a little more help than just this?

 

[PhanthomJay]

I don't understand why I don't need the angle. Can I get a little more help than just this?

You can get the angle from trig, but you have assumed incorrectly that the beam's weight acts at the free of the beam.

 

[PierceJ]

And therefore you should probably correct me right?

 

[axmls]

My apologies. Yes, you need the angle in order to determine the x- and y- components of the tension force.

You have a tension force, the weight, and the two forces from the hinge I mentioned acting on the beam. Now write equations for the x and y net forces.

 

[PierceJ]

Is it
F_x = Tcosθ
F_y = Tsinθ
?

 

[axmls]

Yes, now write the equations for the net force in the x direction, the net force in the y direction, and the net torque.

 

[PierceJ]

Tsinθ(2.41) - mg(1.45) = 0
I don't know what to do with F_x and F_y, I don't think I'm properly understanding.

 

[axmls]

That's almost the net torque. You're forgetting the things I mentioned in the hint from earlier.

 

[PierceJ]

Wait I think I got it here.
Tsinθ(2.41) - mg(1.45) = 0
F_x = Tcosθ
F_y = Tsinθ - mg

 

[axmls]

I'm sorry, I misread something up there. To clear things up:

Your first formula, the formula for the torque, is correct.

Your net x and y force equations are incorrect. The hinge produces a vertical and horizontal reaction force that you need to factor into your equation. Notice that you need to find these for part 2.

 

[PierceJ]

But I just used those and got the right answer?

 

[axmls]

You can get your torque from what you have now. You're supposed to factor in the reaction forces from the hinge to solve this, but I can see what steps would have led to the correct answer.

If you left your equations as they are and just plugged your numbers in, you would get the reaction forces. The "correct" way to solve is to put the reaction forces into the equations, set them all equal to 0, then solve. I imagine it worked for you because of the simplicity of this system.

 

[PierceJ]

It worked on a similar problem that had a weight handing on the end too. We don't do anything more complicated than that so it should work. I mean the parts of the question are split so I have to find the tension first.

 

[haruspex]

you need the angle in order to determine the x- and y- components of the tension force.

You were right the first time. You almost never need to find the angle, as such. Only cos and sin of the angle are needed, and these can be found directly from the linear dimensions.

But I just used those and got the right answer?

Since the reaction at the hinge and the horizontal component of the tension all act through the hinge, they produce no moment about the hinge. But they must be taken into account in $\Sigma F_x$ and $\Sigma F_y$.