Question

A 73 kg adult sits at one end of a 10.5 m long board. His 33 kg child sits on the other end. (a) Where should the pivot be placed so that the board is balanced, ignoring the board's mass? Incorrect: Your answer is incorrect. . m (measured from the adult) (b) Find the pivot point if the board is uniform and has a mass of 15 kg. m (measured from the adult)

Answer #1

Answer

Let x be the distance from the adult to the pivot and then 9.5-x is
the distance to the child.

Now sum torques about the pivot point 73*9.8*x - 33*9.8*(10.5 -x) =
0

Note the 9.8 (g) can be divided out leaving 73x - 33*(10.5-x) = 0
or 106x = 346.5

x = 3.27m (from the man)

b) Now all we do is add the additional torque of the beam (again
eliminating g since it occurs in each term we get

73*x - 15*(10.5/2 - x) - 33*(10.5 - x) = 0

So 121*x = 15*10.5/2 + 33*10.5 = 425.25

So x = 3.51m

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