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:
[tex]F_ed_e=F_ld_l[/tex]
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:
[tex]F_l=\frac{F_ed_e}{d_l}[/tex]
The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.
[tex]F_l=\frac{25*60}{33}[/tex]
F_l=45.5 lbs
