A 2.18-m-long diving board of mass 13.2kg is 3.00m above the water. It has two a

Question

A 2.18-m-long diving board of mass 13.2kg is 3.00m above the water. It has two attahcments holding it in place. One is located at the very back end of the board, and other is 25.0 cm away from the end.

(a) Assuming that the board has a uniform density, find the force acting on each attachment? Answer given ( 435 Newtons upward, 565 Newtons, downward)

(b) If a diver of mass 69.3kg is standing on the front end, what are the forces acting on the two attachments? Answer given (5680 Newtons upward, 6490 Newtons downward)

Explanation / Answer

vertical direction

Fnet = 0

-F1 - M*g + F2 = 0

Net torque = 0 about the back end

M*g(L/2) = F2*x

13.2*9.81*(2.18/2) = F2*0.25

F2 = 564.58 = 565 N

F1 = 435.64 N = 435 N

b) vertical direction

Fnet = 0

-F1 - M*g - m*g + F2 = 0

Net torque = 0 about the back end

M*g(L/2) = F2*x

13.2*9.81*(2.18/2) + m*g*L = F2*0.25

(13.2*9.81*(2.18/2)) + (69.3*9.8*2.18) = F2*0.25

F2 = 6486.1104 N

F1 = 5677.6104 N

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