Evaluate the motion of the system:
If in above Example, the block A is resting on an inclined plane at 45o to horizontal as illustrative in Figure, find out the motion of the system assuming μ = 0.1.
Solution
Supposing that block A moves upwards a distance s while B moves downwards by (s/2) as before as shown in Figure
N A = WA cos 45o
= 400 × 0.707
= 283 N.
Friction = 0.1 × 283
By using the principle of conservation of energy & letting datum level as B′ which is the final position of B,
WA (H + (s/2))+ WB× (s/2) = WA ( H + (s /2)+ s × sin α + (1/2)M A V 2 +(1/2)MB (V/2)2 + 28 × s
∴ 10 × g ( s/2) - 40 × 0.707 s - 28 × s = V 2 (40 +(10/4))
By using equation of motion of uniform acceleration a,
V 2 = 2 a s (since initial velocity is zero)
∴ approximately 49 - 28 - 28 = a (42.5)
∴ a = - 7 /42.5 m / sec2
Negative sign indicates that assumed direction of motion of A is not right and therefore the block A moves downward. As direction of friction also changes (that means upwards), the magnitude of acceleration is recomputed separately.
The system while A moves down through distance s, its velocity is V while Block B moves up with velocity (V/2) considering initial- position of Block B as datum level,
V 2 = 2 (a × s)
Where, a = acceleration of A,
WA (H) = W [H - s (sin α)] + WB (s/2) + 40(V2/2) + (10/2) (V2/4) + F × s
∴ 0 = - 40 × (s × 0.707) × 9.8 + 10 × (9.8 × s)/2 + (V2 /2) (42.5) + 28 s
∴ 0 = s [(- 280 + 49 + 28) + a × 42.5]
Taking 277.14 ≈ 280
∴ a = 203 /42.5
≈ 4.7 m / sec2.
Hence, Block A moves down with acceleration of 4.7 m/sec2.