- #1
Dave1128
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Hi,
I have a question that has been bugging me for a while. I was on a ship where the vessel was rolling from left to right. On board the ship was a sled with angled sides of ~ 60°. the mass of the sled looked to be about 50 Tonnes and the only thing stopping the sled from moving around were some chocks around the base which were shaped parallel to the sides of the sled i.e. ~60° (chamfered chock 40mm x 40mm). I would say the acceleration at the peak of each roll (if that's the right terminology) was about 1.5g i.e. as the ship begins to right itself the sled if forced into the chock at 1.5g. the gap between the sled an chock was about 2mm so insignificant.
My question is what forces were at work to stop the sled from just riding over the chocks and causing havoc. Discounting friction?
Fnorm = 50,00kg * 9.81 = 490,500N
Fgrav = Fnorm
Fapp =50,000 Kg * 1.5g = 735,759 N
Does the angle of the chock exert a force at 60° vertical? how would I calculate this? are there any other forces at play?
Regards
Dave
I have a question that has been bugging me for a while. I was on a ship where the vessel was rolling from left to right. On board the ship was a sled with angled sides of ~ 60°. the mass of the sled looked to be about 50 Tonnes and the only thing stopping the sled from moving around were some chocks around the base which were shaped parallel to the sides of the sled i.e. ~60° (chamfered chock 40mm x 40mm). I would say the acceleration at the peak of each roll (if that's the right terminology) was about 1.5g i.e. as the ship begins to right itself the sled if forced into the chock at 1.5g. the gap between the sled an chock was about 2mm so insignificant.
My question is what forces were at work to stop the sled from just riding over the chocks and causing havoc. Discounting friction?
Fnorm = 50,00kg * 9.81 = 490,500N
Fgrav = Fnorm
Fapp =50,000 Kg * 1.5g = 735,759 N
Does the angle of the chock exert a force at 60° vertical? how would I calculate this? are there any other forces at play?
Regards
Dave