Determine the frictional resistance:
(i) A train having weight 400 kN is running up an inclined path of 1 in 100 at a steady speed of 54 km/hour. If the frictional resistance is 0.5% of its weight, Find out the H. P. (output) of its engine.
(ii) If the steam is cut off whereas the train is ascending this gradient, how far it shall go up the plane before coming to rest. Determine the frictional resistance to remain constant during its travel.
Solution
Figure illustrated the idealized body of train on the gradient (∠ α) moving up the plane under the driving force P.
Frictional force F is, thus, down the plane.
F = (0.5/100) × 400 = 2 kN.
Component of weight W along with the plane is W sin α ≈ 400 ×(1/100)=4KN
Gradient or slope = tan α =1/100 , α being very small and sin α ≈ tam α.
Therefore the driving force P = F + W sin α = 6 kN.
Velocity of train = 54 km/hr
V = 54 × 1000/3600
= 15 m / sec.
(i) ∴ H. P. exerted = P × V/746 = (6000 × 15)/ 746 =120.64
(ii) While the steam is cut off, P = 0
Retarding force = F + W sin α
= 6 kN.
Retardation = a
W = 400 kN
∴ W a /g= 6
or Retardation = 6 × 9.8/400 = 0.147 m / sec^{2} .
When the train comes to rest, V_{2} = 0
(V _{2})^{2} = (V _{1})^{2} - 2 a s
0 = 15^{2} - 2 × 0.147 s
∴ Distance (s) travelled up the gradient = 765 metres.
N. B. Part (ii) of the example may also be solved by using energy principle as follows :
W. D. against resistance = KE available with the train
∴ P × s = W V^{2} /2 g
∴ 6 s = 400 × 15^{2} /2 × 9.8
or,
s = 765 metres