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: A honeybucket man is carrying his load. He has a pole 2 m long with a bucket hanging from each end. The buckets have a mass of 2 kg each. The pole has a mass of 1 kg. The front bucket has 5 kg of "honey" in it, and the rear bucket has 3 kg of "honey." How far from the center of the pole is the honeybucket man's shoulder

Respuesta :

Answer:

Explanation:

Using the principle of moment, assuming the rod is uniform rod of mass 1 kg

the center of mass of the rod will be at 1 m

assuming the system is in equilibrium,

clockwise moment = anticlockwise moment

let the distance of the man shoulder be x from the center of gravity and also is the pivot point

total mass of bucket + mass of honey = 2kg + 3 kg =  5 kg for rear bucket and

2kg + 5 kg = 7 kg for front bucket

( 5kg × ( 1+x)) + ( 1 kg × x) = 7 kg × ( 1 - x)

5 + 5 x + x = 7 - 7x

5 + 6x = 7 - 7x

6x + 7x = 7 - 5

13x = 2

x = 2 / 13 = 0.154 m

the honeybucket man's shoulder is 0.154 m from the center of the pole ( forward ).    

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