Question

Mohon bantuannya.

Jangan lupa caranya.

Answer 1

m1 = 6 kg, m2= 5 kg, m3 = 4 kg, m4 = 3 kg, m5 = 2 kg
F = 40 N
μk = 0,1
g = 10 m/s^2

∑F = m tot . a
F - (fk1 + fk2 + fk3 + fk4 + fk5) = (m1 + m2 + m3 + m4 + m5) . a
F - μk . g (m1 + m2 + m3 + m4 + m5) = (m1 + m2 + m3 + m4 + m5) . a
40 - 0,1 . 10 . (6 + 5 + 4 + 3 + 2) = (6 + 5 + 4 + 3 + 2) . a
40 - 20 = 20 a
20 = 20 a
a = 1 m/s^2

∑F = m5 . a
T54 - fk5 = m5 . a
T54 - 0,1 . 20 = 2 . 1
T54 = 4 N

∑F = m4 . a
T34 - T45 - fk4 = m4 . a
T34 - 4 - 0,1 . 30 = 3 . 1
T34 = 10 N

∑F = m3 . a
T23 - T43 - fk3 = m3 . a
T23 - 10 - 0,1 . 40 = 4 . 1
T23 = 18 N
jadi:
Tegangan tali penghubung benda 2 kg dan 3 kg adalah 4 N
Tegangan tali penghubung benda 4 kg dan 5 kg adalah 18 N