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Question A 5 kg rock is suspended by a massless string from one end of a 7 m measuring stick. 0 1 2 3 4 5 6 7 5 kg What is the weight of the measuring stick if it is balanced by a support force at the 1 m mark? The acceleration of gravity is 9.81 m/s Answer in units of N @ 2019 Colleg
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0 1m 1 2 3 4 5 6 7 2.5m 5 kg mg

The weight of measuring stick is acting vertically downwards at 3.5 m from 0m (2.5m from support)

Take moment about support

sum M=0

5kg * q * 1m_mq*2.5m=0

5q*1-W 2.50

5g- W2.5

5 *9.81/2.5-W

ANSWER:tr = 19.62N

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