Lesson_Text_2.2 - 1 Lesson 2.2 Connected Masses 1 Motion of...

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1 Lesson 2.2 Connected Masses 1. Motion of connected masses . m 1 m 2 m g 2 m g 1 T T Fig. 1. Two connected masses. Fig. 1 on the left shows two masses m 1 and m 2 connected by a string that passes over a pulley. If m 2 is made slightly greater than m 1 , there will be a net force on the system of the two masses and hence m 2 will fall and m 1 will rise with the same acceleration a . When a mass is hanging from a string, the string applies an upward force to hold the mass in place. This is called the tension ( T ) on the string. Our aim here is to obtain the values of a and T . For this let us consider the forces acting on m 1 and m 2 separately. (i) The forces on m 1 : The weight of m 1 , m 1 g acts vertically down and the tension T, the force applied by the string on the mass, acts vertically up. The net force on m 1 , therefore, equal to T – m 1 g. This net force on m 1 produces an acceleration a upward on it which can be represented by the equation: T – m 1 g = m 1 a ……(i) (ii) The forces on m 2 : The weight of m 2 , m 2 g acts vertically down and the tension T acts vertically up. The net force m 2 g – T produces an acceleration a on m 2 downward given by the equation: m 2 g – T = m 2 a …….(ii) Solving equations (i) and (ii) gives g m m m m a 2 1 1 2 + - = …….(iii)
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2 g m m m m T 2 1 2 1 2 + = …….(iv) Example 1 : In fig.7 above, if m 1 = 100 g and m 2 = 101 g, find the acceleration of the masses and the tension on the string. Solution: Taking g = 9.8 m/s 2 , we have: 2 2 1 1 2 0.101 0.100 9.8 0.05 . 0.101 0.100 m m a g m s m m - - - = = × = + + 1 2 1 2 2 2 0.101 0.10 9.8 0.98 0.101 0.10 m m T g N m m = = × = + + In the case we discussed above, both masses are hanging freely. In fig. 8 below, the mass m 1 is placed on a frictionless inclined plane. Here the weight of m 1 , which is m 1 g acts vertically down. Part of this weight is used to pull the mass down the plane and the other part is used to press the object on to the inclined plane. The component of m 1 g down the plane is m 1 g sin θ . m 1 1 m 2 T T m 1 g m 1 g sin θ m 1 g cos θ F n Fig. 2 θ In this case, the net force on m 1 is T m 1 g sin θ . Therefore, T– m 1 g sin θ = m 1 a …1 The net force on m 2 is m 2 g – T, and this gives: m 2 g – T = m 2 a. …2 These two equations may be solved to obtain the values of a and T. Example 2 : In fig. 2 above, if m 1 = 2.25 kg, m 2 = 1.65 kg and θ = 37 o , what will be the acceleration of the system of masses? Solution:
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3 The two equations connecting a and T are: T- m 1 g sin θ = m 1 a m 2 g – T = m 2 a. Solving these two equations for a, we get: 2 1 1 2 2 sin 1.65 9.8 2.25 9.8 sin37 2.25 0.74 1 . .65 m g m g a m m m s θ - - - = = + + = Example 3 : Suppose that θ = 30 o and m 2 = 2.5 kg for the arrangement shown in fig. 8. What should the value of m 1 be if it is to move with an acceleration of (a) 0.25 m/s 2 down the plane, (b) 0.1 m/s 2 up the plane? Solution: (a) The equation for acceleration obtained in example 2 above is for m 1 moving up the plane. When m 1 moves down the plane, the net force on it is m 1 g sin θ – T. The net force on m 2 will then be T – m 2 g. Therefore, the two equations now become m 1 g sin θ - T = m 1 a T – m 2 g = m 2 a
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