Engineering Mechanics

(Joyce) #1

(^638) „„„„„ A Textbook of Engineering Mechanics
Pulls on either side of the rope
We know that pull in the rope with 15 kg mass,
P 1 =m 1 (g – a) = 15 (9.8 – 3.82) = 89.7 N Ans.
and pull in the rope with 5 kg mass,
P 2 =m 2 (a + g) = 5 (3.82 + 9.8) = 68.1 N Ans.
Example 31.11. Two bodies A and B of masses 800 kg and 600 kg are attached at the ends
of a flexible rope. The rope passes over a pulley of 800 mm diameter. The pulley has a mass of
100 kg with a radius of gyration as 400 mm about its axis of rotation.
Find the torque, which must be applied to the pulley to raise the 800 kg body with an accelera-
tion of 1 m/s^2. Neglect friction of the spindle.
Solution. Given: Mass of the body A (m 1 ) = 800 kg; Mass of the body B (m 2 ) = 600 kg;
Diameter of pulley = 800 mm = 0.8 m or radius (r) = 0.4 m; Mass of the pulley (M) = 100 kg; Radius
of gyration (k) = 400 mm = 0.4 m and *acceleration (a) = – 1.0 m/s^2.
We know that pull in the rope carrying 800 kg mass,
P 1 =m 1 (g – a) = 800 [9.8 – (–1)] = 800 × 10.8 = 8640 N
and pull in the rope carrying 600 kg mass,
P 2 =m 2 (a + g) = 600 [(–1) + 9.8] = 600 × 8.8 = 5280 N
We also know that moment of inertia of the pulley,
I=Mk^2 = 100 (0.4)^2 = 16 kg-m^2
and angular acceleration of the pulley,
α=^2
–1
–2.5rad/s
0.4
a
r


∴Torque due to rotation of the pulley,
T 1 =Iα = 16 (– 2.5) = – 40 N-m
and torque acting on the pulley due to difference of pulls in the two ropes,
T 2 =(P 1 – P 2 ) r = (8640 – 5280) 0.4 = 1344 N-m
∴ Torque which must be applied to the pulley
=T 2 – T 1 = 1344 – (– 40) = 1384 N-m Ans.
Example 31.12. A train loaded with cars has a total mass of 6 t is hauled up an incline of
1 in 30 by a rope coiled on a winding drum at the top of the incline. The drum is 1.0 m in diameter,
and has a mass of 1t and has a radius of gyration of 0.45 m. The drum axle friction is negligible.
If a torque of 3 kN-m is applied to the drum from a driving motor, determine the tension in the
rope.
Solution. Given: Mass of train and cars (M) = 6 t; Inclination (sin θ) = 1/30; Diameter
of drum (d) = 1 m or radius (r) = 0.5 m; Mass of drawn (m) = 1 t, Radius of gyration (k) = 0.45 m
and torque applied (T) = 3 kN-m.



  • Under normal circumstances, the body of mass 800 kg is supposed to come down (as it is heavier than
    the other body). Therefore its downward motion is taken as positive. In this case, the body of mass 800
    kg is being raised. Therefore its acceleration is taken as negative.

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