Question

An epicyclic gear train has 3 shafts A, B and C. A is the input shaft running at 100 r.p.m. clockwise. B is the output shaft running at 250 r.p.m. clockwise. The torque on A is 50 kN m (clockwise), C is a fixed shaft. The toque needed to fix C is
1. 20 kN m (anti-clockwise)
2. 20 kN m (clockwise)
3. 30 kN m (anti-clockwise)
4. 30 kN m (clockwise)

Answer 1

Correct Answer - Option 3 : 30 kN m (anti-clockwise)

The net torque applied to the gear train must be zero

T₁ + T₂ + T₃ = 0

The net kinetic energy dissipated by the gear train must be zero.

T₁ω₁ + T₂ω₂ + T₃ω₂ = 0

Here N_A = 100 rpm (cw) ; N_B = 250 rpm (cω); T_A = 50 kN.m (cw) ; N_c = 0

T_AN_A + T_BN_B + T_CN_C = 0

50 × 100 + T_B × 250 = 0

\({T_B} = - \frac{{50 \times 100}}{{250}} = - 20\;kNm\)

T_C = - T_A – T_B = -50 - (-20) = -30 kN.m

i.e 30 kN.m ccw