The uniform diving board has a mass of 35 kg . A B 0.8 m 4 m Find the force on the support A when a 65 kg diver stands at the end of the diving board. The acceleration of gravity is 9.81 m/s 2 . Answer in units of kN.

Respuesta :

Answer:

In other to get the force at the support A, we apply the principle of forces in equilibrium

For a system of forces to be at equilibrium, the sum of all upward forces must be equal to the sum of all downward forces

what are the forces in this system;

for exerted by diver = 65X9.81 = 637.65N = 0.64kN (acting downward)

weight of diving board = 35X9.81 = 343.35N = 0.3434kN (acting downward)

the reaction at point A, the support = RkN (acting upward)

The principle is stated mathematically as:

sum of all upward forces = sum of all downward forces

R = 0.64kN + 0.3434kN = 0.9834kN

hence, the force on the support A is 0.9834kN

Explanation:

First to convert the masses in kg to forces, you multiply by acceleration due to gravity

F = mg

That is how the forces were obtained

Secondly, note that all weight acts downwards and the only force to balance the weights in this system will be the upward reaction from the support at point A

Now you can now apply the principle of forces in equilibrium as stated above to get the unknown force.

Ver imagen ledumbirah

The force on the support A for the uniform diving board is 3.185 kN.

The given parameters;

  • mass of the board, m = 35 kg
  • mass of the diver, = 65 kg

The sketch of the board is presented as follows;

  |-------0.8 m ----|-------------- 4m ----------------|

-Δ-------------------Δ------------------------------------

 A                      B                                       ↓

                                                                65 kg

Take moment about the support B;

[tex]F_A (0.8) = (65 \times 9.8)(4)\\\\F_A (0.8) = 2548\\\\F_A = \frac{2548}{0.8} \\\\F_A = 3185 \ N\\\\F_A = 3.185 \ kN[/tex]

Thus, the force on the support A for the uniform diving board is 3.185 kN.

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