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

AI Thread Summary
The discussion focuses on calculating the downward force exerted by a boy's hand while he balances a sack on a stick. The sack weighs 7.00 kg and is positioned 1.20 m from his shoulder, with his hand located 0.350 m from the shoulder. Using the net torque equation, the force is calculated as 24 N. The conversion from kilograms to newtons is implied through the multiplication by the acceleration due to gravity. The key takeaway is the importance of balancing forces and torques in static equilibrium situations.
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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