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a diving board of length 3.00 m is supported at a point 1.00 m from the end, and a diver weighing 450 n stands at the free end (figure 1). the diving board is of uniform cross section and weighs 260 n. a) Find the magnitude of the force at the support point?
b) Find the direction of the force at the support point? (upward or downward)
c) Find the magnitude of the force at the left-hand end?
d) Find the direction of the force at the left-hand end. (upward or downward)

Sagot :

varshamittal029

a)The magnitude of force at the support point is 1455 N.

b)The direction of the force at the supporting point is in equilibrium.

c)The magnitude of force at the left-hand end is 1030 N.

d)The direction of force at the left-hand end is equilibrium.

Given,

length of board = 3m

distance of pivot = 1.0m

weight of diver = 450N

weight of diving board = 260N

a)

The force at the support point is the sum of all forces multiplied by their distance.

[tex]F_s=f_1d_1+f_2d_2\\[/tex]

here,

f1 = force applied by diver=450 N

d1=distance covered by diver i.e. 1.5 m(half the distance of diving board, because as diver it standing at the centre of board/centre of gravity)

f2=force applied by diving board =260 N

d2=distance covered by diving board = 3m

Now,

[tex]F_s=(450*1.5)+(260*3)\\\\F_s=675+780=1455 N[/tex]

c)

The total torque is zero then,

[tex]f_ld_l-f_1d_1-f_2d_2=0[/tex]

here,[tex]d_1 =1.5-1=0.5m,\d_2=2\ and\ d_l=1m[/tex]

[tex]f_ld_l=f_1d_1+f_2d_2\\\\f_l=(260*0.5)+(450*2)=1030N[/tex]

b) and d)

Because the diver exerts force on both sides at the free end, there is no net force exerted on an object, and it is said to be in equilibrium .Thus, the direction of acting force acting is balanced.

To learn more about equilibrium force refer here

https://brainly.com/question/25329636

#SPJ4

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