Benda bermassa dan dihubungkan dengan tali melalui katrol licin. Jima m1=2 kg m2 =3 kg g=10ms-² maka besar gaya tegangan tali?

ΣF = m₁a
T – m₁g = m₁a
a = (T – m₁g) / m₁

ΣF = m₂a
m₂g – T = m₂a
a = (m₂g – T) / m₂

(T – m₁g) / m₁ = (m₂g – T) / m₂
m₂T – m₁m₂g = m₁m₂g – m₁T
m₂T + m₁T = m₁m₂g + m₁m₂g
(m₁ + m₂)T = 2m₁m₂g
T = 2m₁m₂g / (m₁ + m₂)

T = 2(2)(3)(10) / (2 + 3) = 24 N