Engineering Mechanics

Chapter 25 : Motion of Connected Bodies „„„„„ 525

wards), frictional force equal to 16.97 N (upwards, as the block is moving downwards). Therefore
resultant force

= TB + mB g sin α – 16.97

= TB + 5 × 9.8 sin 30º – 16.97

= TB + 49 × 0.5 – 16.97 = TB + 7.53 ...(v)

Since the body is moving with an acceleration of (a), therefore force acting on it

= 5a ...(vi)

Equating equations (v) and (vi),

TB + 7.53 = 5 a

or TB = 5 a – 7.53 ...(vii)

Now consider the motion of the body C, which is coming down. We know that forces acting on
it are mC.g = 15 × 9.8 = 147 N (downwards) and (TA + TB) upwards. As the body is moving down-
wards, therefore resultant force

= 147 – (TA + TB) ...(viii)

Since the body is moving with an acceleration (a), therefore force acting on it

= 15 a ...(ix)

Equating equations (viii) and (ix),

147 – (TA + TB) = 15 a

Substituting the values of TA and TB from equations (iv) and (vii)

147 – [(4a + 15.68) + (5a – 7.53)] = 15a

147 – 4a – 15.68 – 5a + 7.53 = 15a

∴ 24 a = 138.85

138.85 5.8 m/s 2

24

a== Ans.

(ii) Tension in the two strings

Substituting the value of (a) in equation (iv),

TA = 4a + 15.68 = (4 × 5.8) + 15.68 = 38.88 N Ans.

Again substituting the value of (a) in equation (vii),

TB = 5a – 7.53 = (5 × 5.8) – 7.53 = 21.47 N Ans.