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A father lifts his child as shown in Figure 9.42. What force should the upper leg muscle exert to lift the child at a constant speed?
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<b>Figure 9.42</b> A child being lifted by a father’s lower leg.
Figure 9.42 A child being lifted by a father’s lower leg.
Question by OpenStax is licensed under CC BY 4.0.
$2250\textrm{ N}$
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OpenStax College Physics Solution, Chapter 9, Problem 34 (Problems & Exercises) (1:51)

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This is College Physics Answers with Shaun Dychko. A father is going to make their child rise up at constant velocity on the end of their leg and that will require the counter-clockwise torque equaling the clockwise torque in order to have a dynamic equilibrium. So we are going to have the torque due to the quadricep which will be the force due to the quadricep multiplied by the perpendicular distance to the pivot—which is here— so that's 2.0 centimeters. And then that counter-clockwise torque is gonna be balanced by the two clockwise torques one due to the weight of the leg and that gets multiplied by its perpendicular distance to the pivot—which is 20.0 centimeters— and then add to that the torque due to the child so the child's weight multiplied by the child's perpendicular distance to the pivot—which is 38 centimeters. So we'll solve this for <i>F Q</i> by dividing both sides by <i>r perpendicular</i> and so we have then that the force due to the quadricep then is acceleration due to gravity times mass of the leg times the leg's lever arm— <i>r prime perpendicular</i>— plus the mass of the child multiplied by the child lever arm— <i>r double prime perpendicular— all divided by <i>r perpendicular</i> the lever arm of the quadricep muscle. So that's 9.8 newtons per kilogram times 4.0 kilograms times 20.0 centimeters plus 10 kilograms times 38.0 centimeters all divided by 2.0 centimeters and it's okay to leave your units of centimeters here instead of changing it to meters since there will be centimeters in the top of this fraction and centimeters in the bottom of this fraction which will cancel and we are left with 2250 newtons is the force required from the quadricep muscle.