Find the tension in each rope torque

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SUMMARY

The discussion focuses on calculating the tension in three ropes supporting a uniform plank of length 2m and mass 30kg, with a 700N person positioned 0.5m from the left end. The solution involves isolating the X and Y components of forces, applying the principle of static equilibrium, and using torque equations. The tension in the ropes is derived by setting the sum of moments about the pivot to zero, leading to specific equations for T1, T2, and T3 based on the weights and angles involved.

PREREQUISITES
  • Understanding of static equilibrium and torque principles
  • Knowledge of vector decomposition in physics
  • Familiarity with trigonometric functions, particularly sine and cosine
  • Ability to set up and solve equations based on forces and moments
NEXT STEPS
  • Study the principles of static equilibrium in detail
  • Learn how to apply torque equations in complex systems
  • Explore vector decomposition techniques in physics
  • Practice solving similar statics problems involving multiple forces and angles
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Students in physics, engineers working with statics, and anyone involved in structural analysis or mechanics will benefit from this discussion.

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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