Question

ملح

he crew on Peg Leg Pete the Pirate’s ship have mutinied. Now Pete finds himself walking the plank. The plank is 5.50 m long and has a mass of 16 kg. 3.60 meters of the plank extend out over the shark infested waters of the bay. The plank pivots on a small rail at the edge of the ship (3.60 m from the end of the plank that is over the water); this is the only point on the plank in direct contact with the ship. The remainder of the plank is held in place over the deck by metal bar. The bar is 1.20 m from the pivot. If Pete has a mass of 121 kg, what is the torque provided by the bar when Pete is at the end of the plank. (10 pts)

Answer 1

Mass of Mass 3.6m 1.gm o.fm P Fь of the plank m= 16 16 kg Length of the plank L = 5.5m of the pete M=121 kg Distance from the pivot to FBD of plank Pete in d. = 3.6m Distance from the pivot to centre of the mass of mg my the plank da=0.85m Distance from the pivot to metal bar suppoft dz=1.2m for the equilli brium Eło (about pivot point) Exp = 7, tłz t2z=0 (mg x d.) + (mg xdz) + (23) = Tuz = - (mgxd.) - (mgxdz) 23 (121x9.8* 3.6)-(16*9.880-85) 2.75m 0.857 -D Tz -4268.88 133.28 -4402.16 Nom Luz Porque provided by the bar is hz=-4402-16 Nam -vel sign indicaley torque is clockwise and opposite to the torque provided by plank and rete which is considered to be the. Tz= Fb xdz - F6x1.2 4402.16 Fb = -3668.47N Gorce extested by the bar is Fb = 3668.47N