Static Equilibrium-Balancing Forces and Torque

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SUMMARY

The discussion focuses on calculating the torque applied by a teeter-totter with two children of masses 45 kg and 30 kg. The teeter-totter is 4.0 m long and pivots at its center. The torque formula used is T = r x F sin(theta), where T is torque, r is the distance from the pivot, and F is the force due to gravity. The example provided calculates the torque for the 30 kg child as 147 N·m, using a distance of 1.0 m from the pivot, which requires clarification regarding the positioning of the other child.

PREREQUISITES
  • Understanding of torque and its calculation using the formula T = r x F sin(theta)
  • Knowledge of gravitational force, specifically 9.8 N/kg
  • Familiarity with the concept of static equilibrium in physics
  • Basic understanding of mass and distance in relation to pivot points
NEXT STEPS
  • Study the principles of static equilibrium in physics
  • Learn more about torque calculations in different scenarios
  • Explore the effects of varying mass and distance on torque
  • Investigate the implications of pivot points in mechanical systems
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of torque and equilibrium in practical applications.

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Homework Statement


Two children of masses 45 kg and 30 kg are playing on a teeter-totter of length 4.0m and mass 30.0kg, pivoted at its centre. The heavier child sits 1.75 m from the centre of the teeter-totter that is 0.50 m from the ground. a) What torque does the teeter-totter apply to each of its sides? With this knowledge, how could you simplify any further calculations.

Homework Equations



T=r x Fsin(theta)

The Attempt at a Solution


The example says:
T=rF sin (theta)
T=(1.0 m)(30.0 kg)(9.8 N/kg)
t-t  147 N·m

I don't understand where the one metre is coming from.
Can anyone help me?

 
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The question is too vague to answer. It doesn't specify the coordinates of the other child.
 

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