Problems involving pulleys

1. a ms-2 m kg M kg Problems involving pulleys The shows two particles joined by a light inextensible string which passes over a fixed pulley. If the two particles are of different masses then the heavier particle will move vertically downwards and the lighter particle will move vertically upwards. However, the motions of the two particles are not independent because they are connected by the string.

2. a ms-2 a ms-2 T N T N 3 g N 5 g N Example 1 Particles of masses 5 kg and 3 kg are attached to the ends of a light inextensible string which passes over a smooth fixed pulley. The system is released from rest. Find the acceleration of the system and the tension in the string and also find distance moved by the 5 kg mass in the first 2 seconds of the motion. (Assume that neither particle reaches the pulley or the ground.) Solution T – 3g = 3a (1) 5g – T = 5a (2) (1) + (2) gives : 2g = 8a So a = 2.45 ms-2 Sub the value of a into (1) Gives T = 36.75 N u = 0 t = 2 a = 2.45 s = ? s = ut + ½ at2 s = 0 x 2 + 0.5 x 2.45 x 22 So s = 4.9 m

3. R N a ms-2 T N P 15 N a ms-2 T N 6 g N Q 4g N Example 2 Two particles P and Q of masses 6 kg and 4 kg respectively are connected by a light inextensible string. Particle P rests on a smooth horizontal table. The string passes over a smooth pulley fixed at the edge of the table and Q hangs vertically. There is a friction of 15 N opposing the motion of the particle. The system is released from rest. Find (a) the acceleration (b) the tension in the string and (c) the force exerted on the pulley.

4. R N a ms-2 T N P 15 N a ms-2 T N 6 g N 29.52 Q 4g N 29.52 Solution For the 4 kg mass: 4g – T = 4a [1] For the 6 kg mass: T – 15 = 6a [2] [1] + [2] gives: a = 2.42 ms-2 From [2]: T = 29.52 N Force exerted on the pulley: Resultant force = (29.522 + 29.522) = 41.7 N Acting at 45º to the horizontal

5. R a T2 T1 4 kg T1 4 g N T2 10 kg 5 kg 10 g N 5 g N Example 3 A block of mass 4 kg is attached by light, inextensible strings to particles of mass 5 kg and 10 kg. The string pass over smooth pulleys at either end of a smooth horizontal able, as shown. Find the acceleration of the system and the tensions in the two strings.

6. R a T2 T1 4 kg T1 4 g N T2 10 kg 5 kg 10 g N 5 g N Solution For the 10 kg mass: 10g – T1 = 10a [1] For the 4 kg mass: T1 – T2 = 4a [2] For the 4 kg mass: T2 – 5g = 5a [3] 10g - T2 = 14a [4] [1] + [2] gives: 5g = 19a [3] + [4] gives: Gives: a = 2.58 ms-2 [1] Gives: T1 = 72.2 N [3] Gives: T2 = 61.9 N

7. R N a ms-2 T N P F N a ms-2 T N 8g N Q 4g N Example 2 Two particles P and Q of masses 8 kg and 4 kg respectively are connected by a light inextensible string. Particle P rests on a rough horizontal table. The string passes over a smooth pulley fixed at the edge of the table and Q hangs vertically. The coefficient of friction between P and the table is 0.3. The system is released from rest. Find (a) the acceleration (b) the tension in the string and (c) the force exerted on the pulley.

8. R N a ms-2 T N P F N a ms-2 T N 8g N Q 4g N Solution (a) Resolve (P  ) R – 8g = 0 R = 8g F = R = 0.3 x 8g = 2.4g Resolve (P ) T – F = 8a T – 2.4g = 8a (1) Resolve (Q  ) 4g - T = 4a (2) (1) + (2) 1.6g = 12a gives a = 1.31 ms-2 (b) T – 2.4g = 8a gives T = 34 N

9. 34 N 34 N 34 N  34 N Resultant Solution (c) Resultant = (342 + 342) = 48.1 N at 450 below horizontal

10. T T 5 kg 3g N 300 Example 3 A body of mass 5 kg is placed on a smooth plane inclined at 300 to the horizontal. It is connected by a light inextensible string, which passes over a smooth pulley at the top of the plane, to a mass of 3 kg hanging freely. Find the common acceleration and the tension in the string (1) T – 5gsin300 = 5a : T – 2.5g = 5a 3g - T = 3a (2) (1) + (2) : 0.5g = 8a so a = 0.6125 T – 2.5g = 5a So T = 27.6 N