An ocean-going barge is 50.0 m long and 20.0 m wide and has a mass of 145 metric ton. Will the barge clear a reef 1.50 m below the surface of the water?

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anonymous · Answer

Summing forces we get that:$$F _{net}=0=F _{barge}-F _{buoyant}$$Now we the force the barge exerts is:$$F _{barge}=mg$$where m is the mass of the barge and g is the acceleration of gravity.  Fbuoyant is given by:$$F _{buoyant}=\rho _{w}Vg$$where ρw is the density of water; and V is the volume of water displaced.  We can substitute these last two equations into the first to get:$$mg-\rho _{w}Vg=0$$$$mg=\rho _{w}Vg$$And if we simplify we get:$$m=\rho _{w}V$$Now V is given by:$$V=lwd$$where l is the length of the barge; w is the width of the barge; and d is the depth to which the barge sinks.  Now we have:$$m=\rho _{w}lwd$$Now all you need to do is solve for d.