How much force is required to lift a 200 lb weight with a class 1 lever, given the effort arm is 8 feet and the resistance arm is 2 feet?

To determine the amount of force required to lift a 200 lb weight using a class 1 lever, we can utilize the principle of leverage, which states that the force applied (effort) is inversely proportional to the lengths of the effort arm and the resistance arm.

In a class 1 lever, the formula that relates effort (E), load (L), effort arm (EA), and resistance arm (RA) can be expressed as:

[ E \times EA = L \times RA ]

where:

• E is the effort force,

• EA is the length of the effort arm,

• L is the load (or the weight to be lifted), and

• RA is the length of the resistance arm.

Given:

• Load (L) = 200 lbs,

• Effort arm (EA) = 8 feet,

• Resistance arm (RA) = 2 feet.

We can rearrange the formula to solve for the effort:

[ E = \frac{L \times RA}{EA} ]

Substituting the values into this equation:

[ E = \frac{200 , \text{lbs} \times 2 , \text{feet}}{8 , \text{feet