Static Equilibrium Question; Please Help.

Question

A person is standing on a diving board that is supported by two springs, one at each end. The diving board weighs 200 N and is 2.0 m long. What is the force exerted by each spring when a 500-N person stands 0.5 m from one end?

Answer 1

Let

F₁ be the unknown upward force of one spring,

F₂ be the unknown upward force of the other string,

B = -200 N be the downward weight of the board,

P = -500 N be the downward weight of the person,

L = 2.0 m be the length of the board,

r = 0.5 m be the distance of the person from the end 1 with force F₁.

I am using the usual convention that upward direction is positive and downward direction is negative.

Therefore the forces B and P are negative.

The forces F₁ and F₂ will turn out to be positive.

Since the board is in static equilibrium the following must hold:

0) The net force must be 0

F₁ + F₂ + B + P = 0

1) The net torque around the first point must be 0

F₂L + BL/2 + Pr = 0

The equation is the sum of the torques of all four forces.

The force F₁ does not appear because its distance from point 1 is 0.

The weight B of the board can be considered all concentrated at the

center of gravity, which is distance L/2 from point 1.

2) The net torque around the second point must be 0

F₁L + BL/2 + P(L-r) = 0

The equation is the sum of the torques of all four forces.

The force F₂ does not appear because its distance from point 2 is 0.

So we have three equations and only two unknowns, F₁ and F₂.

That is OK, because the equations are not linearly independent.

We will use equation 1) to calculate F₂,

use equation 2) to calculate F₁, and then

use equation 0) just to double check the results.

F₁ = -(BL/2 + P(L-r))/L = -B/2 - P + Pr/L

F₂ = -(BL/2 + Pr)/L = -B/2 - Pr/L

If we substitute the solutions for F₁ and F₂ into equation 0) we indeed get 0.

Substituting actual numbers:

F₁ = 200/2 + 500 - 500×0.5/2 = 475 N

F₂ = 200/2 + 500×0.5/2 = 225 N