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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 Fn = 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» \ pillu 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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This note was uploaded on 10/13/2010 for the course PHYSICS 101 taught by Professor Abdulrazzaq during the Spring '10 term at West Virginia State University.
 Spring '10
 Abdulrazzaq
 Physics

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