Question

A lightweight diving board is supported by two pillars. One is at the end of the board and the other is 1.55 m away. A 75-kg person is standing on the board at such a location that the pillar at the end (pillar #1) of the board exerts a 810-N force downward. Determine the following.

(a) distance the person is standing from the end pillar (pillar #1)

(b) magnitude of the upward force exerted by the second pillar (pillar #2)

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the person = M = 75 kg

Downward force exerted by the first pillar on the diving board = F₁ = 810 N

Upward force exerted by the second pillar on the diving board = F₂

Distance between the two pillars = d = 1.55 m

Distance of the person from the first pillar = d₁

Distance of the person from the second pillar = d₂

Taking moment balance about the second pillar,

F₁d = Mgd₂

(810)(1.55) = (75)(9.81)d₂

d₂ = 1.706 m

d₁ = d + d₂

d₁ = 1.55 + 1.706

d₁ = 3.256 m

Balancing the forces on the diving board,

F₂ = F₁ + Mg

F₂ = 810 + (75)(9.81)

F₂ = 1545.75 N

a) Distance of the person from the end pillar (pillar #1) = 3.256 m

b) Magnitude of the upward force exerted by the second pillar = 1545.75 N