Find the tension in each rope torque

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Homework Help Overview

The discussion revolves around a statics problem involving a uniform plank supported by three ropes, with a person applying a load. Participants are tasked with finding the tension in each rope while considering the effects of torque and equilibrium.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss isolating variables and breaking down forces acting on the system. There are questions about the placement of the ropes and the pivot point for calculating moments. Some participants suggest treating the person as a point load and considering the weight of the plank at its center of mass.

Discussion Status

The discussion has progressed with some participants offering insights into the setup of equations for moments and forces. One participant reports having figured out a method to calculate tensions, indicating that some productive direction has been achieved, though no consensus on the final solution has been reached.

Contextual Notes

Participants note difficulties with the diagram's clarity and the exact positioning of the ropes, which may affect the interpretation of the problem. There is also mention of treating the system under the assumption of no tension in the plank.

poloboy
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Find the tension in each rope "torque"

1. A uniform plank of length 2m and mass 30kg is supported by three ropes. Find the tension in each rope when 700N person is 0.5m from the left end.

2. diagram
↑ /
| /
| /40 degrees ↓ (rotating clockwise)
←-----person-----------------/---


The Attempt at a Solution



So i began with isolating the X and Y variables
ie: X on left, fcos40
y going up, fsin40
how would i go about breaking down the mass of the person + plank to figure out the tension in each rope?
 
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For a statics problem the sum of the moments about any pivot is zero. It's a little difficult to tell from the illustration where the ropes are, but I'm guessing all three are attached at the left end?

Treat the person as a point load from at 0.5 m from the end, or 1.5 m from the pivot. Treat the weight of the plank at its center of mass.
 
sorry the diagram didn't show up better

there's a rope pointing left, one pointing up (on the left side) and one rope pointing tot he right at 40 degrees North east
 
And the pivot is on the right?

If there is no tension in the plank then then the x- or horizontal component in the rope at 40° will balance the tension in the horizontal rope.

Try to write the equation for the sum of the moments about the pivot.
 
awsome, figured it out.
i set x coord t1cosO - T3 and y coord as T1sinO + T2 - W(person+plank)

set my t1 = 1/4(Wperson +2Mplank)/sinO then i just plugged in T1 to my other forumlas.

thanks!
 

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