How Does the Boy Balance the Forces While Carrying the Sack?

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SUMMARY

The discussion centers on calculating the downward force exerted by a boy's hand while balancing a sack on a stick. The sack has a mass of 7.00 kg and is positioned 1.20 m from the shoulder, while the hand is 0.350 m from the shoulder. Using the principle of net torque equating to zero, the calculated force is 24 N, derived from the formula: (mass * distance from shoulder) / (distance from shoulder to hand).

PREREQUISITES
  • Understanding of basic physics concepts such as torque and equilibrium
  • Familiarity with Newton's second law of motion
  • Knowledge of unit conversion from kilograms to Newtons
  • Ability to perform algebraic calculations involving distances and forces
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  • Learn about the conversion of mass to weight using gravitational acceleration
  • Explore real-world applications of torque in engineering and mechanics
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lacar213
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Homework Statement


A boy carries a sack on one end of a very light stick that is balanced on his shoulder, at an angle of 22.0 degrees up from the horizontal. The mass of the sack is 7.00 kg, and it sits 1.20 m from his shoulder. If his hand is 0.350 m from his shoulder, on the other end of the stick, what is the magnitude of the downward force the boy exerts with his hand?


Homework Equations


net torque = 0


The Attempt at a Solution


7.00 * 1.20 / .350 = 24 N
 
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lacar213 said:

Homework Statement


A boy carries a sack on one end of a very light stick that is balanced on his shoulder, at an angle of 22.0 degrees up from the horizontal. The mass of the sack is 7.00 kg, and it sits 1.20 m from his shoulder. If his hand is 0.350 m from his shoulder, on the other end of the stick, what is the magnitude of the downward force the boy exerts with his hand?

Homework Equations


net torque = 0

The Attempt at a Solution


7.00 * 1.20 / .350 = 24 N

And you got from kg to N how?
 

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