How Do You Calculate Tension and Pole Force on a Leaning Tether Ball?

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SUMMARY

The discussion focuses on calculating the tension in the rope and the force exerted by the pole on a tether ball leaning against it. Given a string length of 1.80m, a ball radius of 0.200m, and a mass of 0.500kg, the tension in the rope is calculated to be 4.93N. The force exerted by the pole on the ball is determined to be approximately 0.548N. The calculations utilize static equilibrium equations and trigonometric functions to derive these values.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Knowledge of basic trigonometry, specifically sine and cosine functions
  • Familiarity with Newton's second law of motion
  • Ability to perform calculations involving forces and angles
NEXT STEPS
  • Review static equilibrium problems in physics
  • Study the application of trigonometric functions in force calculations
  • Learn about tension in ropes and cables in various contexts
  • Explore real-world applications of forces acting on objects in equilibrium
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the mechanics of forces acting on objects in equilibrium, particularly in practical scenarios like tether ball systems.

Edwardo_Elric
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Homework Statement


A tether ball leans against the post to which it is attached. If the string to which the ball is attached is 1.80m long, the ball has a radius of .200m, and the ball has a mass of .500kg, what are the tension in the rope and the force the pole exerts on the ball? assume that there is so little friction between the ball and the pole that it can be neglected. ( The string is attached to the ball such that a line along the string passes through the center of the ball.)
EquilibriumProblem.jpg




Homework Equations


[tex]F_{x} = 0[/tex]
[tex]F_{y} = 0[/tex]


The Attempt at a Solution


Part I
Required: tension on the rope
T = ??
wball = (.500kg)(9.8m/s^2) = 4.9N
[tex]\sum{F_{y}} = 0 = T\cos{\theta} + (-w_{ball})[/tex]
[tex]T = \frac{4.9N}{\cos{\theta}}[/tex]
*I computed theta using its distances:
sin(theta) = x_{radius} / h_{tension}
= arcsin {0.200m/1.80m}
= 6.38 degress
[tex]T = \frac{4.9N}{\cos{6.38}}[/tex]

T =4.93N

Part II
reqd: force on pole on ball:
F_{pole on ball} = ?
\sum{F_{x}} = 0 = T\sin{\theta} - F_{p.on.B}
(4.93N)sin{6.38} = F_{pole on ball}
F = .548N?
 
Last edited:
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