1.70. '1 V-21/(3W+ kg).
1.71. (a) w^1 — ("^11 — "^1 2) g-1-^2 m2wo
± m2(b) F = rn 4 :+ 1 7: 1 (g— w^0 ).
1.72. w= 2g (2i — sina)/(41 + 1).1.73. (^) = 4mini 2 -1-mo (m1— ins)
(^1) 4m1m2+ mo + m2) 5.
1.74. Fir = 21mMI(M — m) t^2.
1.75. t =1/2/ (4 ± Ti)/3g (2— r() 1.4 s.
1.76. H = /(1-1 + 4) = 0.6 m.
= ml (^7) 17m2 2 (g w0):
Fig. 4. Fig. 5.
1.77. W A = gl(1 cote a), w B = g/(tan a + cot a).
1.78. w= g V D(2+ k+
1.79. wmia = g (1 — k)/(1 k).
1.80. wmax = g (1 k cot a)/(cot a — k).
1.81. w = g sin a cos a/(sin^2 a^ ml/m^2 ).
1.82. w —
mg sin a
M 2m (1— cos a) '
1.83. (a) l(F)1= 2 -1/ 2 mv2htR; (b) l(F)1= mac.
1.84. 2.1, 0.7 and 1.5 kN.
1.85. (a) w V1+ 3 cos 2 0, T =3mg cos 0;
(b) T= mg j/3; (c) cos 0=1/V "g,^ = 54.7°.
1.86. 53°.
1.87. 0 = arccos (2/3) ^ 48°, v = V 2gR/3.
1.88. a = 1/(x/nuo 2 — 1). Is independent of the rotation di-
rection.
1.89. r =R/2, vmax = 1 /2 kgR.
1.90. s = 1 / 2 R V (kg/w.0^2 —1 = 60 m.
1.91. v C a V kg/a.
1.92. T = (cot 0 + co 2 RIg) mg/2n.
1.93. (a) Let us examine a small element of the thread in contact
with the pulley (Fig. 5). Since the element is weightless, dT =
dF j,. = k dF„ and dF„ = T da. Hence, dTIT = k da. Integrat-