an 8 N weight is placed at one end of a meterstick. a 10N weight is placed at the other end. where should the fulcrum be placed to have the meterstick balanced?

Explanation:

= 8 N

= 10 N

Since the meter stick has a length of 1 m, and

Let = x

Let = 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:

where: w - weight

d - distance from the fulcrum

Substituting the value of and 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 =

x = 0.56

= x

= 0.56 m

= 1 m - x

= 1 m - 0.56 m

= 0.44 m

1

explanation: