A small ferryboat is 4.00 m wide and 6.00 m long. When a loaded truck pulls onto it, the boat sinks an additional 4.30 cm into the water. What is the weight of the truck? In N?
2006-11-08 07:07:10 · 3 answers · asked by activegirl 1 in Science & Mathematics ➔ Physics
10124 N
1032 Kg
2006-11-08 07:15:14 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The two answerers above know the solution. I'll try here to give you the steps and the calculations to solve the problem.
Based on Archimedes' principle, the weight of the volume of water displaced by the ferryboat when the truck pulls into it is equal to the bouyant force on the boat. Now that bouyant force is equal to the weight of the truck. Why? Because that is necessary to maintain the equilibrium of the boat. Check Newton's law as well, ie: for every action there is an equal and opposite reaction.
Therefore, bouyant force,
bf={[4*6*4.30/100]*1000}9.8
Note the density of water is 1000kg/m^3, and to convert the mass to N, we multiply by a factor equal to 9.8. We also converted cm to m by dividing 4.30cm by 100.
If the ferryboat is on sea water then change the density to 1030kg/m^3. Since it is only a small ferryboat, I assumed it is on plain water not on sea water.
bf=10113.6N
weight of truck=10,113.6N
2006-11-08 19:03:28 · answer #2 · answered by tul b 3 · 0⤊ 0⤋
Calculate the volume of the additional water displaced. Multiply by the density of water (sea water is 1.03*10^3 Kg/meter^3). That's the truck's mass. W= mg.
2006-11-08 07:44:22 · answer #3 · answered by sojsail 7 · 0⤊ 0⤋