Explanation:
[tex] w_1 [/tex] = 8 N
[tex] w_2 [/tex] = 10 N
Since the meter stick has a length of 1 m, and [tex] d_1 \:+\:d_2\:=\:1\:m [/tex]
Let [tex] d_1 [/tex] = x
Let [tex] d_2 [/tex] = 1 - x
Where should the fulcrum be placed to have the meterstick balanced?
For the system to be balanced, the product of the weight and distance of the objects on opposite sides should be equal. This is is shown by the equation:
[tex]w_1 d_1 = w_2 d_2 [/tex]
where: w - weight
d - distance from the fulcrum
Substituting the value of [tex] w_1, \:w_2, \:d_1 \:[/tex] and[tex] \: d_2[/tex] in the formula,
(8 N)(x) = (10 N)(1 m - x)
x(8 N) = (10 N)m - x(10 N)
x(8 N) + x(10 N) = (10 N) m
x(18 N) = 10 N
x = [tex] \frac{(10 \:N)m}{18 \:N}[/tex]
x = 0.56
[tex] d_1 [/tex] = x
[tex] d_1 [/tex] = 0.56 m
[tex] d_2 [/tex] = 1 m - x
[tex] d_2 [/tex] = 1 m - 0.56 m
[tex] d_2 [/tex] = 0.44 m
