The pressure applied to a leverage bar varies inversely as the distance from the object. If 150 pounds is required for a distance of 10 inches from the object how much pressure is needed for a distance of 3 inches

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Answer:

500 pounds

Step-by-step explanation:

Let the pressure applied to the leverage bar be represented by p

Let the distance from the object be represented by d.

The pressure applied to a leverage bar varies inversely as the distance from the object.

Written mathematically, we have:

[tex]p \propto \dfrac{1}{d}[/tex]

Introducing the constant of proportionality

[tex]p = \dfrac{k}{d}[/tex]

If 150 pounds is required for a distance of 10 inches from the object

  • p=150 pounds
  • d=10 inches

[tex]150 = \dfrac{k}{10}\\\\k=1500[/tex]

Therefore, the relationship between p and d is:

[tex]p = \dfrac{1500}{d}[/tex]

When d=3 Inches

[tex]p = \dfrac{1500}{3}\\\implies p=500$ pounds[/tex]

The pressure applied when the distance is 3 inches is 500 pounds.