1.70. '1 V-21/(3W+ kg).
1.71. (a) w1 — ("11 — " 12) g-1- 2m2wo
± m2
(b) F = rn4: 1
+ 17: (g— w0).
1.72. w= 2g (2i — sina)/(41 + 1).
1.73.
= 4mini2 -1-mo (m1— ins)
1 4m1m2+ mo
+ m2) 5.
1.74. Fir = 21mMI(M — m) t2.
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. WA = 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/(sin2a
ml/m2).
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 cos2 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/nuo2 — 1). Is independent of the rotation direction.
1.89. r =R/2, vmax = 1/2 kgR.
1.90. s = 1/2R V (kg/w.02 —1 = 60 m.
1.91. v C a V kg/a.
1.92. T = (cot 0 + co2RIg) 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-