838 PM edugen.wileyplus.com Assignment Gradebook ORION Downloadable eTextbook nm

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838 PM edugen.wileyplus.com Assignment Gradebook ORION Downloadable eTextbook nment MESSAGE MY INSTRUCTOR FULL SCREEN PRINTER VERSION BACK Chapter 09, Problem 12 The figure shows a person whose weight is w 688 N doing push-ups. Find the normal force exerted by the floor on (a) each hand and (b) each foot, assuming that the person holds this position. (a) Number (b) Number Units TO TEXT LINK TO By accessing this Question Assistance, you will learn while you earm points based on the Point Potential Policy set by your Question Attempts: 0 of 4 used E TE -I SAVE FOR LATER e Earn Maximum Points available only if you answer this question correctly in three attempts or less. d@ 6 7 8 9

Explanation / Answer

Solution: If the person holds the position, then there is no acceleration nor rotation, so we have vertical and rotational equilibrium.

Vertical equilibrium tells us all vertical forces are balanced, which means
2F+2H = 688 N; where F is he force of the floor on each foot and H is the force of the floor on each hand

Thus F + H = 344 N..........Equation 1

the sum of torques is also zero; if we sum torques around the feet, we have
sum of torques = -688 L1 + 2H*(L1+L2) =0 .............Equation 2; where L1=0.84m, L2=0.41m

Note that the force F exerts no torque around the feet since F has no lever arm around the feet.

substituting values in Equation 2 , we have

-688*0.84 + 2H*(0.84+0.41) =0

Thus, -577.92 + 2.5H = 0 ; Thus H = 231.168 N (Force on each Hand) Answer (a)

Thus, from equation 1, F + H = 344 N

This implies F + 231.168 N = 344 N. Thus, F = 112.832 N (Force on each Foot) Answer (b)

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