Engineering Mechanics

(^518) „„„„„ A Textbook of Engineering Mechanics
Equating equations (iv) and (v),
147 – 2T = 7.5 a ...(vi)
Adding equations (iii) and (vi),
10.4 = 27.5 a
or a =
10.4
27.5
= 0.4 m/s^2
∴ Velocity of the body A after 10 seconds, if the system starts from rest.
v = u + at = 0 + 0.4 × 10 = 4 m/s Ans.
Example 25.10. The system of bodies shown in Fig. 25.13 starts from rest.
Fig. 25.13.
Determine the acceleration of body B and the tension in the string supporting body A.
Solution. Given : Weight of body A (W 1 ) = 500 N ; Weight of body B (W 2 ) = 750 N and
coefficient of friction (μ) = 0.2
Acceleration of the body B
T = Tension in the strings, and
a = Acceleration of the body B.
From the slope of the surface, we find that
tan α =
3
4
= 0.75 or sin α = 0.6 and cos α = 0.8
We know that normal reaction on the inclined surface due to body of weight 750 N
= W 2 cos α = 750 × 0.8 = 600 N
∴ Frictional force =μR = μ × 600 = 0.2 × 600 = 120 N
First of all, consider the motion of the body B, which is moving, upwards. We know that the
forces acting on it, along the plane, are 2T newtons (upwards), W 2 sin α newtons (downwards) and
force of friction equal to 120 newtons (downwards as the body is moving upwards). Therefore
resultant force
= 2T – W 2 sin α – 120 = 2T – 750 × 0.6 – 120 N
= 2T – 570 N ...(i)
Since the body is moving with an acceleration (a), therefore force acting on it
2
2
750
76.5
9.8
W
mg a a a
g
=×= ×= ...(ii)
Equating equations (i) and (ii),
2 T – 570 = 76.5 a ...(iii)