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## Homework Statement

The board's left end is fastened to a fixed support by a connection that is free to pivot. The board is supported to its middle on another fixed support point. A diver of mass 60 kg stands at the end of the board. (ignore the mass of the board for this problem)

The board extends

**2.2 m**from the pivot.

The diver is

**60 kg**.

I) Suppose the board deflects by 3 cm when the 60 kg diver stands on it. Now he jumps up and comes down on the board; at the lowest point of his motion, the deflection of the board is 6 cm. What is the net force on the diver?

II)A new board is installed that is twice as thick as the old board but is made of the same material and is the same length and width. Approximately how much does this board deflect when the 60 kg diver stands on it?

## Homework Equations

[tex]\vec{F}_{sp} = k \Delta x[/tex]

[tex]\Delta L = \frac{FL}{AY}[/tex]

[tex]\Delta x = \frac{\Vert \vec{F} \Vert}{k}[/tex]

thus,

[tex]k = \frac{AY}{L}[/tex]

## The Attempt at a Solution

I solved part I.

[tex]k = \frac{\vec{F}_{sp}}{\Delta x}[/tex]

[tex]k = \frac{60 \cdot 9.8}{.03} = 19600 N/m[/tex]

[tex]\vec{F}_{sp} = k \Delta x = 19600 \cdot .06 = 1176 N[/tex]

I'm having a problem with part II. I figure since the area is doubled that I would use Young's modulus (Y) to solve this. But the following formula needs and area (A) and I don't know what to put for it.

[tex]k = \frac{AY}{L} = 1176 = \underbrace{\frac{AY}{2.2}}_\text{two unknowns}[/tex]