Calculating Tension in a Revolving Mass: Homework Solution

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SUMMARY

The discussion focuses on calculating the tension in two wires supporting a 970g mass revolving horizontally at a speed of 7.5 m/s. Using the equations of motion, the tension in the upper wire (T1) is determined to be 46 N, while the tension in the lower wire (T2) is calculated to be 26.868 N. The forces in both the vertical and horizontal directions are analyzed, leading to the conclusion that T1 and T2 can be derived from the equations of equilibrium and centripetal force.

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  • Understanding of Newton's laws of motion
  • Familiarity with centripetal force concepts
  • Knowledge of trigonometric functions, specifically cosine
  • Ability to solve simultaneous equations
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  • Study the application of Newton's laws in circular motion
  • Learn about tension in multi-wire systems
  • Explore the derivation of centripetal acceleration formulas
  • Investigate the role of angles in force calculations
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Students in physics, particularly those studying mechanics, engineers working with tension systems, and anyone involved in circular motion dynamics.

sunnnystrong
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Homework Statement



The figure shows two wires that are tied to a 970g mass which revolves in a horizontal circle at a constant 7.5 m/s. What is the tension in the upper wire?

The wires are both 1m long, and are attached to a pole. The radius is cos(30 degrees). It's an equilateral triangle.

Homework Equations



∑F = m*a

The Attempt at a Solution


[/B]
∑Fy = 0.
∑Fx = (m*v^2) / r

First I looked at the forces in the y-direction. Down = negative.
The upper rope = T1. The lower rope = T2.

(.5)T1 - (.5)T2 - mg = 0
*Solve for T1*
T2 + 2*m*g = T1 // EQN 1

Next, in the X-Direction. Inward = positive.

T1cos(30) + T2cos(30) = (m*v^2)/r
Plug in value for T1 to solve for T2.
2*T2*cos(30) + 2*m*g*cos(30) = (m*v^2)/r
[(m*v^2)/r - 2*m*g*cos(30) ] / (2cos(30) = T2
T2 = 26.868 N

Plug into EQN 1 and I got T1 = 46 N
 
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Looks good.
 
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gneill said:
Looks good.
Thank you!
 

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