Question

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

Answer 1

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:

$p \propto \dfrac{1}{d}$

Introducing the constant of proportionality

$p = \dfrac{k}{d}$

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

• p=150 pounds
• d=10 inches

$150 = \dfrac{k}{10}\\\\k=1500$

Therefore, the relationship between p and d is:

$p = \dfrac{1500}{d}$

When d=3 Inches

$p = \dfrac{1500}{3}\\\implies p=500 pounds$

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