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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?

Sagot :

HannahPandaidn

The fulcrum should be placed 0.44 from the 12 N weight or 0.56 m from the 8 N weight.

Explanation:

Given:

[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

Question:

Where should the fulcrum be placed to have the meterstick balanced?

Equation:

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

Solution:

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

Solve for [tex] d_1 [/tex]

[tex] d_1 [/tex] = x

[tex] d_1 [/tex] = 0.56 m

Solve for [tex] d_2 [/tex]

[tex] d_2 [/tex] = 1 m - x

[tex] d_2 [/tex] = 1 m - 0.56 m

[tex] d_2 [/tex] = 0.44 m

Final Answer:

The fulcrum should be placed 0.44 from the 12 N weight or 0.56 m from the 8 N weight.

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