Chap 6 - 24, Chap 8 - 13, 14 (7th edition)

Chap 6 - 24, Chap 8 - 13, 14 (7th edition) - (a) B exerts a...

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Unformatted text preview: (a) B exerts a horizontal force on A. (b) A exerts a force on B that is opposite in direction to the force B exerts on A. (c) The force on A is equal in magnitude to the force on B, but is oppositely directed. (d) Yes. The momentum of the system (the two skaters) is conserved because the net external force on the system is zero (neglecting friction). (e) (AprWm =(Ap,)d+(ap,)3=0 :9 mA(vA—O)+m3(vB—0)=0 or “(mg ( M l a}, _ LE vs =—LWJ(2.00 m/s] =—2.22 m/s VA = 2.22 m/s in the direction opposite to $3 In each case, the distance from the bar to the center of mass of the body is ' zm'x' _ mamsxams + mmrsoxtor‘so + mthighs xthighs + miegsxiegs cg 2:rni mams + mmrso + mthfghs + miegs where the distance x for any body part is the distance from the bar to the gravity of that body part. I each a - w .. 1tak the u c '_ - ir m to run frgm the bar toward the lggag'gn Qf the head. Note that: Emj = [6.87 + 33.57 +14.07 + 7.54) kg = 62.05 kg With the body positioned as shown in Figure P8.13b, the distances at for mill". is computed using the sketch given below: P Positive directit Kflee Hip Shoulder i," 10th Joint Joint *‘3 thighs "(r )' W ) Cg I$85 53' thighs xm = +(r;g )m = +0239 m xm = + (r = +0548 m+0.337 m = 0.885 m as )mrso xmgm = 8m + Em + (ng )Wgh = (+0.548+0.601+0.151) m = 1.30 m xkgs = em + + 8W + [rcg Legs = [+ 0.548+0.601+0.374+0.227) m. - +62. k - . With these distances and the given masses we find: xC : 8 g it? 8 62.05 kg With the body positioned as shown in Figure P8.13c, we use the folio determine the distance x for each body part: ' Positive direction Hip Shoulder Knee Joint Joint Rotational Equifibrtum and Rotational Dynamics 289| x = +(rcg)m = +0239 111 = 1.1,”,15—(123)t = +0548 m —0.337 m = +0211 m xwgm = — gram—(reg)ng = (+0548 — 0.601 — 0.151] m = “0.204 m x, = E J 0W, — 11W — (r )W = [+ 0.548 — 0.601 — 0.374 ~ 0.227) m = —0.654 m age arms cg 1th these distances, the location (falative *0 the bar) 0f the center 0f graVity 0f the bOdY .924 k - acts = fig = + 0.015 m = 0.015 111 towards the head '7“; the coordinate system shown below, the coordinates of the center of gravity of 1-: body part may be computed: Bar 1+? \: (FUEL W N48" 8 arms ‘ [r63 )thighs 600° +x origin/WI em #(r )m = 0.309 m K D ycgw arms cg Ir )mso = m ycg, torso = 0 mm +{r )W h 00560.00 = 0.676 m ycglihighs = (reg)W sin60.0° = 0.131 m ' lg s I I5 ‘8 "rm + 0W cos 600° + (ch )m = 1.02 m 31%,835 = ems,“ sin 60.00 = 0.324 m ' ' coordinates for individual body parts and the masses given in Problem 8.13, mam: + mtorso + mrhi'ghs + mlegs ...
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This note was uploaded on 04/07/2012 for the course PHYS 111 taught by Professor Dr.jackman during the Fall '07 term at Waterloo.

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Chap 6 - 24, Chap 8 - 13, 14 (7th edition) - (a) B exerts a...

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