Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re

Question

3 years ago

11

Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re

sistance to the fulcrum, what is the maximum amount of weight that can be lifted with 25 lbs of effort.
A. 165 Lbs
B. 13.8lbs
C. 45.5lbs
D. 3.79lbs

Physics

1 answer:

leva [86] · Answer 1 · 2021-12-24T02:53:37+03:00

Answer:

The correct answer is C. 45.5 lbs.

Explanation:

In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.

The formula for any problem involving a lever is:

$F_ed_e=F_ld_l$

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.

The parameter of the formula that you need is F_l:

$F_l=\frac{F_ed_e}{d_l}$

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

$F_l=\frac{25*60}{33}$

F_l=45.5 lbs