Suspended 82617

A round steel weight (ρ1=7800 kg/m3) is suspended on a thread and immersed in water (ρ2=1000 kg/m3). The volume of the weight is 1dm3. With what force is the thread stretched?

Correct answer:

F =  68 N

Step-by-step explanation:

ρ1=7800 kg/m3 ρ2=1000 kg/m3 g=10 m/s2 V=1 dm3 m3=1:1000  m3=0.001 m3  m=ρ1 V=7800 0.001=539=7.8 kg  F1=m g=7.8 10=78 N F2=ρ2 V g=1000 0.001 10=10 N  F=F1F2=7810=68 N



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