Question

A diving board of length 3.00 mm is supported at a point 1.00 mm from the end, and a diver weighing 550 NN stands at the free end (Figure 1). The diving board is of uniform cross section and weighs 260 NN.

Answer 1

Force at the supporting point = 1890 N

Force at the left hand end = 1130 N

What is Equilibrium force?

When there is no net force exerted on an object and it is considered to be in equilibrium if the amount and direction of the forces acting on it are precisely balanced.

Calculation of the force:

Given:

Length of diving board = 3 m

Distance of pivot = 1.0 m

$d_{1}$ = 1.5 m distance of board's center of gravity to left end measurement

Weight of the diver = 500 N

Weight of the diving board = 260 N

Calculation of the force at the supporting point:

The total force multiplied by each force's respective distance equals the force at the support point.

$F_{s} = f_{1} d_{1}+f_{2}d_{2}$

$F_{s}$ = 260 x 1.5 + 500 x 3

$F_{s}$ = 1890

supporting force at a distance of 1 m.

$F_{s} = \frac{1890}{1}$ = 1890 N

Hence, the force at the supporting point is 1890 N.

Calculation of force at the left hand end:

Net torque = 0

Therefore, $F_{1}d -f_{1}d_{1}-f_{2} d_{2}=0$

$F_{1} d=f_{1}d_{1} +f_{2} d_{2}$

$d_{1}$ = 1.5 - 1 = 0.5 m

$d_{2}$ = 2 m

d = 1

$F_{1}$ = (260 x 0.5)+(500 x 2)/1

$F_{1}$ = 1130 N

Hence, the force at the left hand end is 1130 N.

Learn more about Equilibrium forces here:

brainly.com/question/25329636

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