Question
Figure 9.24(a) shows a wheelbarrow being lifted by an applied force Fi. If the wheelbarrow is filled with twenty bricks massing 3 kg each, estimate the value of the applied force Fi. Provide an explanation behind the total weight w and any reasoning toward your final answer. Additionally, provide a range of values over which you feel the force could exist.
<b>Figure 9.24(a)</b> (a) In the case of the wheelbarrow, the output force or load is between the pivot and the input force. The pivot is the wheel's axle. Here, the output force is greater than the input force. Thus, a wheelbarrow enables you to lift much heavier loads than you could with your body alone.
Figure 9.24(a) (a) In the case of the wheelbarrow, the output force or load is between the pivot and the input force. The pivot is the wheel's axle. Here, the output force is greater than the input force. Thus, a wheelbarrow enables you to lift much heavier loads than you could with your body alone.
Question by OpenStax is licensed under CC BY 4.0
Final Answer

60 N≤Fi≤180 N60 \textrm{ N} \le F_i \le 180\textrm{ N}

Solution video

OpenStax College Physics for AP® Courses, Chapter 9, Problem 10 (Test Prep for AP® Courses)

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Video Transcript
This is College Physics Answers with Shaun Dychko. Given a wheelbarrow with some bricks inside it, we want to figure out what input force is needed for lifting straight up on the handles here to balance this wheelbarrow and there are 20 bricks and each having a mass of 3 kilograms, we need to make estimates as to what the lever arm of each force is. So we have the weight straight down and we can ignore this output force here— there's just a weight straight down due to weight of the bricks— and the perpendicular distance between that force and the pivot at the center of the wheel I am estimating to be about 0.3 meters— it's about the length of a ruler... it's a regular 12 inch ruler, 30 centimeter ruler— and then an estimate for this distance between where the hands are on the handle and the pivot, let's say is 1.5 meters. We know that the total clockwise torque has to equal the total counterclockwise torque and clockwise speaking, we have this input force multiplied by its lever-arm r i and that's going to equal the total weight which is the number of bricks times the mass for a single brick times g multiplied by the lever-arm for the weight and we divide both sides by r i to solve for F i. So that's 20 bricks times 3 kilograms per brick times 9.8 newtons per kilogram times 0.3 meters— lever-arm for the weight— divided by 1.5 meters lever-arm for the input force that's 120 newtons and these estimates, you know, are not very precise or they are not very certain so we can say there's an error 50 percent plus or minus in which case, the input force would be somewhere between 60 newtons and 180 newtons.