Engineering Mechanics

(^522) „„„„„ A Textbook of Engineering Mechanics
Since the body is moving downwards with an acceleration (a), therefore force acting on it.
= m 2 a = 25 a ...(v)
Equating equations (iv) and (v)
83.8 + T = 25 a ...(vi)
Adding equations (iii) and (vi),
211.1 = 40 a
or
211.1 5.3 m/s 2
40
a==
∴ Distance moved by the bodies in 3 seconds,
()
(^1120) 5.3 3 (^2) 23.85 m
22
sut at=+ =+× = Ans.
25.8.MOTION OF TWO BODIES CONNECTED BY A STRING AND LYING
ON ROUGH INCLINED PLANES
Fig. 25.17.
In this case, the friction between the bodies of masses m 1 and m 2 and the surfaces AB and BC
will also be considered.
It may be noted that the force of the friction always acts in the opposite direction to the motion
of a mass.
If the mass m 1 comes down and the mass m 2 goes up, the frictional force at m 1 will act upwards
and the frictional force at m 2 will act downwards. Thus, while calculating the forces on any mass, the
frictional force, acting in the opposite direction, should also be considered.
Example 25.13. Two rough planes inclined at 30º and 15º to the horizontal and of the same
height are placed back to back. Two bodies of masses of 15 kg and 5 kg are placed on the faces and
connected by a string over the top of the planes If the coefficient of friction be 0.3 find from
fundamentals the resulting acceleration.
Solution. Inclination of first plane (α 1 ) = 30º ; Inclination of second plane (α 2 ) = 15º ; Mass
of first body (m 1 ) = 15 kg ; Mass of second body (m 2 ) = 5 kg and coefficient of friction (μ) = 0.3.
Fig. 25.18.