Question

a diving board of length 3.00 m is supported at a point 1.00 m from the end, and a diver weighing 530 n stands at the free end(figure 1). the diving board is of uniform cross section and weighs 250 n. for related problemsolving tips and strategies, you may want to view a video tutor solution of locating your center of gravity while you work out. part a find the magnitude of the force at the support point. express your answer in newtons. part b find the direction of the force at the support point.

Explanation:

Step1: Identify the torques

Let the support point be the pivot. The weight of the diving board \(W_{board}=250\ N\) acts at its center of mass (at \(L/2 = 1.5\ m\) from the end). The diver's weight \(W_{diver}=530\ N\) acts at the end (\(3 - 1=2\ m\) from the support). Let the force at the support be \(F_{support}\). Taking torques about the support point, \(\tau = rF\sin\theta\). Since all forces are perpendicular to the board (\(\theta = 90^{\circ},\sin\theta = 1\)). The torque due to the board's weight \( \tau_{board}=250\times(1.5 - 1)\) (distance from support is \(1.5 - 1=0.5\ m\)), and the torque due to the diver's weight \(\tau_{diver}=530\times(3 - 1)\)

Step2: Apply the torque - equilibrium condition \(\sum\tau=0\)

Counter - clockwise torques equal clockwise torques. Let the force at the support create a counter - clockwise torque. \(\tau_{support}=\tau_{board}+\tau_{diver}\). \(F_{support}\times0=\tau_{board}+\tau_{diver}-F_{support}\times0\) (using \(\sum\tau = 0\), \(F_{support}\times0\) is the torque about the support due to \(F_{support}\) itself, which is zero). \(F_{support}\times0=(250\times0.5)+(530\times2)\) (wrong approach, actually using \(\sum F_y = 0\): \(F_{support}-W_{board}-W_{diver}=0\) is wrong. Using torque about the end (non - support end) for a better approach. Let's use the correct torque method: Take torques about the non - support end. Let \(x = 1\ m\) (distance from non - support end to support). The weight of the board acts at \(1.5\ m\) from the non - support end, and the support force acts at \(x = 1\ m\) from the non - support end. \(\sum\tau=0\): \(F_{support}\times1-250\times1.5 - 530\times3=0\)

Step3: Solve for \(F_{support}\)

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Answer:

\(1965\ N\)