(^608) A Textbook of Engineering Mechanics
Fig. 30.6. Motion on inclined plane.
Now consider a body moving downwards on an inclined plane as shown in Fig. 30·6 (a).
Let m = Mass of the body, and
α = Inclination of the plane.
We know that normal reaction on the inclined plane due to mass of the body.
R = mg cos α ...(i)
This component is responsible for the force of friction experienced by the body.
We also know that component of the force along the inclined plane due to mass of the body
= mg sin α ..(ii)
This component, responsible for sliding (or moving ) the body downwards, is known as
gravitational pull. Now we can find out work done in moving the body or power required for the
motion by subtracting the force of friction (due to normal reaction) from the component along the
inclined surface (mg sin α).
Note. If the body is moving upwards, then the component along the inclined surface is taken
as an additional resistance i.e. this component is added to other types of resistances.
Example 30.10. Calculate the work done in pulling up a block of mass 200 kg for 10 m on
a smooth plane inclined at an angle of 15° with the horizontal.
Solution. Given : Mass of the block (m) = 200 kg ; Distance (s) = 10 m and inclination of
plane (α) = 15°
We know that resistance due to inclination
= mg sin α = 200 × 9·8 sin 15° = 1960 × 0.2588 = 507.2 N
and work done = Resisting force × Distance = 507·2 × 10 = 5072 N-m/s
= 5.072 kN-m/s = 5.072 kJ Ans.
Exmple 30.11. A locomotive and train together has a mass of 200 t and tractive resistance
100 N per tonne. If the train can move up a grade of 1 in 125 with a constant speed of 28·8 km.p.h.
find the power of the locomotive.
Also find the speed, which the train can attain, on a level track, with the same tractive resis-
tance and power of the locomotive.
Solution. Given : Mass of the body i.e. locomotive and train together (m) = 200 t ; Tractive
resistance = 100 N/t = 100 × 200 = 20 000 N = 20 kN ; Grade
1
sin 0·008
125
α= = and speed of the
train (v) = 28·8 km/h = 8 m/s
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