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-g A diving board 3.00 m long is supported at a point 1.00 m from the end, and a diver weighing 500 N stands at the free end (Fig. E11.11). The diving board is of uniform cross section and weighs 280 N. Find (a) the force at the support point and (b) the force at the left-hand end.

Respuesta :

Answer:

Explanation:

Let the force at the support end is R and at the left end is F.

According to the equilibrium condition

F + R = 280 + 500

F + R = 780 .... (1)

Take moments about G.

F x 1.5 + R x 0.5 = 500 x 1.5

3F + R = 1500 .... (2)

Subtract equation (1) from equation (2), we get

2F = 1500 - 780

F = 360 N

R = 780 - 360

R = 420 N

(a) Force at support is 420 N

(b) force at the left end is 360 N.

Ver imagen Vespertilio

The force at the supporting point is 1920N and the force at the left-hand end is 1140N

Data;

  • length of board = 3m
  • pivot distance = 1.0m
  • weight of diver = 500N
  • weight of diving board = 280N

Force at the Support point

The force at the support point is the summation of all the forces multiplied by their respective distance

[tex]F_s = f_1 d_1 + f_2d_2\\f_s = 280*1.5 + 500*3\\f_s = 1920[/tex]

The supporting force at a 1m distance

[tex]f_s = \frac{1920}{1} = 1920N[/tex]

The force at the supporting point is 1920N

The force at Left-Hand end

This can be calculated as

[tex]F_l d - f_1d_1 - f_2d_2 = 0\\f_l d = f_1d_1 + f_2d_2\\d_1 = 1.50 - 1.0 = 0.50m\\d_2 = 2m\\d = 1\\f_l = \frac{(280*0.50) + (500*2)}{1}\\ f_l = 1140N[/tex]

The force at the left-hand end is 1140N

Learn more on equilibrium forces here;

https://brainly.com/question/25329636

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