Find the Tension: 2-m Rod with .6kg Mass and 2kg Suspended Mass

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Homework Help Overview

The problem involves a 2-meter long uniform rod suspended horizontally by two vertical strings at its ends, with a mass of 0.6 kg and an additional 2 kg mass suspended from the rod. Participants are discussing how to determine the tension in each string while considering the equilibrium of the system.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are exploring the concept of torque and its application to the problem, questioning which masses to include in their torque equations. There is discussion about the direction of torque and how it relates to the equilibrium of the system.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of torque and equilibrium. Some have suggested calculating torques with respect to different points, while others are clarifying the role of the suspended mass in influencing the tensions in the strings.

Contextual Notes

Participants are navigating the complexities of the problem without explicit angles or detailed methods, focusing instead on the relationships between forces and torques in an equilibrium scenario.

catenn
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A 2-m long uniform rod AB is suspended horizontal by two vertical strings attached to the ends A and B. The rod has a mass of .6kg. A mass of 2kg suspended from the rod .8m from the end A. Determine the Tension in each string.

Hi, I have a physics worksheet I am trying to work on. This has confused me, I'm unsure of what to put inside of the equation without any angles and two different strings. I have labeled A T1 and B T2 and I know T1>T2. Also Torque must = 0 for the system to be in equilibrium so I make an equation = to T0 but if anyone could help with what to put into the equation I need some help. Thanks so much!
 
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What to put in the equation? Well, torques, for example. :smile: If you know the definition of a torque, there shouldn't be any problems.
 
Right, but which masses? Is it something like Torque0 = T1(2.6)(.8) - T2(1)(.6)? Are they subtracted or added?
 
catenn said:
Right, but which masses? Is it something like Torque0 = T1(2.6)(.8) - T2(1)(.6)? Are they subtracted or added?

You can start by calculating setting the torque with respect to point A (or B) equal to zero. If the torque 'rotates' clockwise, choose a positive sign, and if it 'rotates' counter clockwise, choose a negative sign.
 
Its not rotating either way, the whole beam is in equilibrium. The two strings are holding it up and I need their tensions.
 
catenn said:
Its not rotating either way, the whole beam is in equilibrium. The two strings are holding it up and I need their tensions.

I know it's not rotating, I didn't mean that literarely. I was talking about the direction of the torque.
 
catenn said:
Its not rotating either way, the whole beam is in equilibrium. The two strings are holding it up and I need their tensions.
What are the tensions notwithstanding the 2 kg mass? What does the 2 kg mass add to the downward force of each end of the bar? How does this influence each tension?
 
It would cause a downward counter clockwise motion that is positive. The tension is greater for the string on A than B w/ more weight. The weights need to be converted to Newtons and multiplied by 9.81 for gravity.
 
catenn said:
It would cause a downward counter clockwise motion that is positive. The tension is greater for the string on A than B w/ more weight. The weights need to be converted to Newtons and multiplied by 9.81 for gravity.
What is the effect of the added mass on each string?
 
  • #10
As said before, use the equations of equilibrium. How is equilibrium expressed? What must vanish? You can use two torque equations (with respect to points A and B), and use the fact that the sum of the vertical forces must vanish as a check.
 

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