Find the force as a function of distance...? A 4.80 m long diving board of negli

Question

Find the force as a function of distance...?

A 4.80 m long diving board of negligible mass is supported by two pillars. One pillar is at the left end of the diving board; the other is 1.70 m away. A 88.0 kg diver stands at the far end of the board.

a) Find F 1 as a function of the distance, x, of the swimmer from the left end of the diving board. Assume that the diving board is uniform and has a mass of 82.0 kg .

b) Find F 2 as a function of the distance x.

e F 2=((0.180kN/m)x+0.532kN)y^

Answer for pt. a is NOT a, pt. b is NOT c

a F 1=((0.508kN/m)+1.14kN)y^ b F 1=((0.180kN/m)x+1.14kN)y^ c F 1=((0.180kN/m)x1.14kN)y^ d F 1=((0.508kN/m)x0.532kN)y^ e F 1=((0.508kN/m)x+0.532kN)y^ d

Explanation / Answer

net torque about the right pillar = 0

F1*d = - [ M*g*(x-d) + m*g*L/2 ]

]F1*1.7 = -[88*9.81*x - 88*9.8*1.7 + 82*9.81*(2.4-1.7)]

F1 = ( -0.508 (kN/m) x + 0.532 kN ) y

option (e)

=================

along vertical

Fnet = 0

F1 + F2 - mg - Mg = 0

F2 - 508x + 532 - (82*9.81) - (88*9.81) = 0

F2 = (+508x + 1140 )y

F2 = ((0.508(KN/m)x + 1.14 KN ) y

OPTION (a)

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