Engineering Mechanics

(^610) „„„„„ A Textbook of Engineering Mechanics
and work done in one second = Total resisting force × Distance
= 101.5 × 10 = 1015 kN-m/s = 1015 kJ/s
∴ Power = 1015 kW Ans.
Example 30.13. A truck of mass 5 tonnes just moves freely without working the engine at
18 km.p.h. down a slope of 1 in 50. The road resistance at this speed is just sufficient to prevent any
acceleration.
Determine the track resistance in newtons per tonne mass of the engine. What power will it
have to exert to run up the same slope at double the speed, when the track resistance remains the
same?
Solution. Given : Mass of the truck (m) = 5t = 5000 kg ; Velocity of the truck v = 18 km.p.h.
= 5 m/s and slope
1
(sin ) 0.02
50
α= =.
Track resistance per tonne mass of truck
We know that resistance due to inclination
= mg sin α = 5000 × 9.8 × 0.02 = 980 N
Since the engine is not working and the truck is moving without any acceleration, therefore no
external force is acting on it. Or in other words, the track resistance is equal to the resistance due to
inclination (i.e. 980 N). Therefore track resistance per tonne mass of the truck
980
196 N
5
= Ans.
Power to be exerted by the engine for moving the truck upwards
Since the truck is moving up the slope, therefore it has to overcome tractive resistance (980 N )
plus resistance due to inclination (980 N).
∴Total resisting force = Tractive resistance + Resistance due to inclination
= 980 + 980 = 1960 N = 1.96 kN
and work done in one second = Total resisting force × Distance
= 1.96 × (2 × 5) = 19.6 kN-m/s = 19.6 kJ/s
∴ Power = 19.6 kW Ans.
Example 30.14. An army truck of mass 5 tonnes has tractive resistance of 150 N/t. Find the
power required to propel the truck at a uniform speed of 36 km.p.h. (a) up an incline of 1 in 100 ; (b)
on a level track ; and (c) down an incline of 1 in 100.
Solution. Given : Mass of the truck (m) = 5 t = 5000 kg ; Tractive resistance = 150 N/t
= 150 × 5 = 750 N ; Speed of the truck (u) = 36 km.p.h = 10 m/s and slope
1
(sin ) 0.01
100
α= =.
(a) Power required to propel the truck up the incline,
Since the truck is propelled up the incline, therefore it has to overcome tractive resistance plus
resistance due to inclination. We know that resistance due to inclination.
= mg sin α = 5000 × 9.8 × 0.01 = 490 N
∴ Total resisting force = Tractive resistance + Resistance due to inclination
= 750 + 490 = 1240 N
and work done in one second = Total resisting force × Distance = 1240 × 10 N-m/s
= 12 400 N-m/s = 12.4 kN-m/s = 12.4 kJ/s
∴ Power = 12.4 kW Ans.