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
[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.
