Radhika Batra
Last Activity: 11 Years ago
Let T1 = Tension in the string tying m1 to m2.
and T2 = Tension in the string tying m2 to m3.
Free body diagram for the masses and corresponding equations are
T1–µkm1g = M1a
T2–T1–µkm2g = m2a
F–T2–µkm3g = m3a
Solving these equations we get
(a) m3a = F–µkg(m1+m2+m3)–m1a–m2a
= 14.02 m/s2
(b) T1 = m1 a + m1 µk g
T1 = 20 (14 + 0.98) = 299.60 N
T2 = 299.60 + 30 × 14.98 = 749 N