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# ch9 - 12 From the force diagram for the mass we can write...

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Unformatted text preview: 12. From the force diagram for the mass we can write 21-“x = 13T1 — FTZ cos 6 = 0, or FT1 = FT2 cos 30°. Hy: Fusin 9— mg = 0, or FT2 sin 30° = mg = (200 k)(9.80 m/sz), which gives Pr. = Thus we have FT] = I:T2 cos 30° = (3.9 x 103 N) cos 30° = 3.4 x103 N. 13. From the force diagram for the hanging light and junction we can write U, = FT1 cos 91 — I:T2 cos 62 = 0; FT1 cos 37° = FT2 cos 53°; 02 a] XFy=FTlsin61+FT2sin62—mg=0; x FT1 sin 37° + FT2 sin 53° = (30 kg)(9.80 m/sz). When we solve these two e uations for the two unknowns, P“, and P“, we get F-n = 1.8 x102 N, and F12 = 2.4 x 102 N. 18. We choose the coordinate system shown, with positive torques clockwise. We write it = let about the support point A from the force diagram for the seesaw and boys: in, = + ngﬂ + mggx — mIggL = 0; + (35 kg)§(3.6 m) + (25 kg)x — (50 kg);(3.6 m) = 0, which gives x = 1.1 m. The third boy should be 1.1 m from pivot on side of lighter boy. 23. We choose the coordinate system shown, with positive torques clockwise. We write 21 = Ia about the point A from the force diagram for the beam: 21,, = — (FT sin a)L + MggL = 0; — FT sin 50° + (30 kg)(9.80 m/sz)§ = 0, Note that we find the torque produced by the tension by finding the torques produced by the components. We write IF = ma from the force diagram for the beam: 2FI = FWX— FT cos a = 0; PM - (1.9 x 102 N) cos 50° = O, which gives FWJr = 123 N. IF}, = Fwy + FT sin a—Mg = 0; Fwy + (1.9 x 102 N) sin 50° — (30 kg)(9.80 m/sz) = 0, which gives Fwy = 147 N. For the magnitude of Fw we have PW = (I-‘Wx2 + Fwy2)‘/2 = [(123 N)2 + (147 N)2]1/2 = 1.9 x 102 N. We find the direction from tan 6: Fwy/wa = (147 N)/(123 N) = 1.19, which gives 0 = 50°. Thus the forcelat the wall is PW = 1.9 x 102 N, 50° above the horizontal. (P p» \ pill-u 26. We choose the coordinate system shown, with positive torques clockwise. We write 21’ = let about the point A from the force diagram for the beam and sign: 21A = — (I:T sin 9)D + MgL + mgﬁL = 0; — I?T (sin 41.0°)(1.35 m) + (215 N)(1.70 m) + (135 N)%(1.70 m) = 0, which gives Note that we find the torque produced by the tension by finding Prairie: the torques produced by the components. A We write 2F = ma from the force diagram for the beam and sign: Ur: FhinseH—FTCOS 6: PhinseH - (542 N) cos 410° = 0, which gives FhinseH = ZFy = Phinsev + FT sin 9— Mg — mg = 0; (542 N) cos 41.0° — 215 N — 135 N = 0, WhiCh gives FhingeV= l:hingeV 409 N. 27. We choose the coordinate system shown, with positive torques clockwise. We write 21’ = Ia about the point A from the force diagram for the pole and light: 21A = - FTH + MgL cos 0 + mgiL cos 9: O; — FT (3.80 m) + (12.0 kg)(9.80 m/sz)(7.5 m) cos 37° + (8.0 kg)(9.80 m/52)s12(7.5 m) cos 37° = 0, We write 2F = ma from the force diagram for the pole and light: z"Far-:FAH—F'I‘:0" FAH— 2.5 x 102 N = 0, which gives PAH: ZFy=FAV—M8'm8=0} FAV— (12.0 kg)(9.80 m/sz) — (8.0 kg)(9.80 m/sz) = 0, which gives PM: 36. We choose the coordinate system shown, with positive torques FM clockwise. We write 21' = Ia about the shoulder joint A from the force diagram for the arm: 9 CC 21A = ng — (PM sin 9)d = 0; A (3.3 kg)(9.80 m/52)(O.24 m) — (PM sin 15°)(O.12 m) = 0, F] L— m8 which gives PM = d <— D ...
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ch9 - 12 From the force diagram for the mass we can write...

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