Question

A diver of mass 48.0kg stands 0.5 m from the end of a uniform diving board over the water. The board is 9.00m long and has a mass of 85kg. The end of the board opposite the diver is attached to the platform. The bounce of the board is adjusted by a roller underneath that can be moved back and forth. The roller is 2.80 m from the fixed end of the board. a) What are the magnitude and the direction of the force on the board due to the roller

Answer 1

The magnitude of the force on the board is 5,566.401 N

The direction of the force acting on the roller is upwards

Reasons:

The given parameters are;

Mass of the diver = 48.0 kg

Mass of the diving board = 85 kg

Distance the diver stands from the end of the diving board = 0.5 m

Length of the diving board = 9.00 m

Distance of the roller from the fixed end of the diving board = 2.80 m

a) Required:

The magnitude and direction of the force on the board due to the roller

Solution:

Taking moments about the roller gives;

∑M = 0

Clockwise moments = Counterclockwise moments

48 × (8.5 - 2.8) × 9.81 + 85 × (4.5 - 2.8) × 9.81 = $F_y$ × 2.8

Where;

$F_{yf}$ = Vertical force at fixed point

• $F_{yf} = \dfrac{4,101.561}{2.8} \approx 1,468.84$

$F_{yf}$  = 1,464.84 N

At equilibrium, the sum of the vertical force is zero, therefore;

$\sum F_{y} = 0$

Which gives;

• $\downarrow \sum F_{y} = \ \uparrow \sum F_{y}$

$\downarrow \sum F_{y}$ = Weight of diver + weight of board + vertical force at fixed end of board

Therefore;

$\downarrow \sum F_{y}$ = 48 × (8.5 - 2.8) × 9.81 + 85 × (4.5 - 2.8) × 9.81 + 1,464.84 = 5566.401

$\downarrow \sum F_{y}$ = 5566.401

• $\uparrow \sum F_{y}$ = Upward force from roller

Therefore;

• Upward force from roller = $\uparrow \sum F_{y}$ = $\downarrow \sum F_{y}$ = 5566.401 N

The magnitude and direction of the force on the board due to the roller is 5,566.401 N, upwards

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