Understanding Tension in Angles: Expert Insights and Tips

[Icefire_16]

Homework Statement

1). What happens to tension as you move the mass closer to Axis of Rotation? Explain the physics of this.
2). What happens to tension as you increase the angle(theta)? Explain the physics of this.
(The L, length of the beam stays constant and is hinged to the wall. The M, constant , mass can move closer or farther from the hinge. Theta, or the angle can be change if the cable is at a higher position or lower.)

Relevant Equations

1). I said it decreases because there is less weight pulling at the cable when in is further in.
2). I said it increases but I am not sure why. Is it because of the steeper angle that the cable becomes tighter?

 

[berkeman]

Welcome to the PF.

Sorry, your post is a bit confusing for me. Move what mass toward what axis of rotation? There appear to be two masses labeled "M" in your diagram...

And is the "axis of rotation" at the intersection of the horizontal M mass and the wall?

And what "angle" where? There are no angles labeled in the diagram. We can try guessing at all of these issues, but it would be best if you could make the problem statement super clear. Thanks!

 

[Icefire_16]

Welcome to the PF.

Sorry, your post is a bit confusing for me. Move what mass toward what axis of rotation? There appear to be two masses labeled "M" in your diagram...

And is the "axis of rotation" at the intersection of the horizontal M mass and the wall?

And what "angle" where? There are no angles labeled in the diagram. We can try guessing at all of these issues, but it would be best if you could make the problem statement super clear. Thanks!

Thanks! Sorry I'm new here but I updated the picture, hopefully it's more clear!

 

[gneill]

So, is the horizontal beam massless? The original picture showed it having a mass M.

 

[Icefire_16]

So, is the horizontal beam massless? The original picture showed it having a mass M.

Whoops sorry it does have a mass

 

[gneill]

Whoops sorry it does have a mass

Is its mass different from the suspended one? In other words, does m = M in your revised diagram?

 

[Icefire_16]

Is its mass different from the suspended one? In other words, does m = M in your revised diagram?

Nope the masses are different

 

[haruspex]

Homework Statement:: 1). What happened to tension as you moved mass closer to Axis of Rotation. Explain the physics of this.
2). What happens to tension as you increased the angle(theta)? Explain the physics of this.

These questions are unclear. You cannot change either of these without changing something else, so we need to know what is being held constant.
E.g. in the first one, if it is mass M that is moving left, is that by making L shorter (so m moves left too), and if so is the angle increasing or is the point of attachment of the wire dropping?

 

[Icefire_16]

These questions are unclear. You cannot change either of these without changing something else, so we need to know what is being held constant.
E.g. in the first one, if it is mass M that is moving left, is that by making L shorter (so m moves left too), and if so is the angle increasing or is the point of attachment of the wire dropping?

Updated!

 

[Cutter Ketch]

Now the question is clear. However, your answers sound like you are guessing.

Given the nature of the problem, I am sure you have been told a couple principles that will allow you to answer these questions with good physics reasoning. What physics principles have you been taught that apply in this situation?

 

[Icefire_16]

Now the question is clear. However, your answers sound like you are guessing.

Given the nature of the problem, I am sure you have been told a couple principles that will allow you to answer these questions with good physics reasoning. What physics principles have you been taught that apply in this situation?

I have been taught torque and static equilibrium but I am kind of struggling to grasp the topic conceptually.

 

[gneill]

If the situation is static (no movement) , then the sum of all torques must be zero. That's the crux of the analysis procedure.