The drawing shows a person (weight W = 594 N, L1 = 0.853 m, L2 = 0.397 m) doing

Question

The drawing shows a person (weight W = 594 N, L1 = 0.853 m, L2 = 0.397 m) doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position.

Find force on each hand N

Find force on each foot N

I understand and saw the solutions:

[0.853/(.397+0/853)] X 594 N = 405.3

*******BUT, why do you have to subtract this number from the W=594????***************

I understand the next step is (594-405.3)/2 but why can't you just divide 405.3 by 2 and have that the answer?

Explanation / Answer

Taking moment about hand:

594*0.397 = Foot(N)*( 0.853+0.397)

Foot(N) = 188.6544 N

Taking moment about leg:

594*0.853 = HandN)*( 0.853+0.397)

Hand(N) = 405.3456 N

Force on each hand =405.3456/2 = 202.6728 N

Force on each foot = 188.6544/2 = 94.3272 N

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