Answer :
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
Final answer:
In this scenario, we are dealing with a second class lever. First, we calculated the mechanical advantage using the given distances. Using this mechanical advantage, we can multiply it with the effort applied to find the maximum weight that can be lifted, which is 45.5 lbs.
Explanation:
The subject of this question is physics within the scope of mechanics, specifically the topic of levers. A second class lever is one in which the resistance (load) is found between the fulcrum (pivot point) and the effort (force applied). In the context of this problem, we derive the mechanical advantage of the lever using the formula: Mechanical Advantage = Effort Distance / Resistance Distance.
To proceed, we need to make sure we're dealing with consistent units. Since the effort distance is given in feet and the resistance distance in inches, we ought to convert the latter to feet: 33 inch = 33/12 feet = 2.75 feet.
Now, inputting the given values, we get: Mechanical Advantage = 5.00 feet / 2.75 feet = 1.82 (rounded to two decimal places). To find the maximum weight that can be lifted with 25 lbs of effort, we multiply the effort applied by the Mechanical Advantage: 1.82 * 25 lbs = 45.5 lbs. Therefore, the answer is (C) 45.5 lbs.
Learn more about Second Class Lever here:
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