Forces and Newton's Laws of Motion 187

(ii) In the situation shown in figure-2.102, (a) What minimum forcewill make any part for whole system move. (b) For the following valuesof force, find the acceleration of two blocks, nature and value of frictionat the two surfaces 2 N and 6 N.

[(a) 4 N, (b) 1 m/s2, 0]

(iii) In previous question if force acts on the lower block, (a) Where does the sliding begins first. (b) Whatis the minimum force at which any part of system starts sliding. (c) At what value of force F will the slidingstarts at the other surfaces. (d) For the following values of F, find the acceleration of the two blocks, natureand value of friction at both rough surfaces -3 N, 12 N, 24 N.

[(a) lower (b) 6 Nt (c) 18 Nt (d) 0 m/s2, 3 Nt, 0 Nt, 1 m/s2, 6 Nt, 2 Nt, 3.5 m/s2, 2 m/s2, 6 Nt, 4 Nt]

(iv) In the situation shown in figure-2.103, (a) for what maximum value of force F can all three blocks movetogether. (b) Find the value of force F at which sliding starts at other rough surfaces. (c) Find accelerationof all blocks, nature and value of friction force for following value of force F - 10 N, 18 N and 25 N.

Figure 2.103

[(a) 15 Nt (b) 21 Nt (c) 2 m/s2, 3 m/s2, 4 m/s2, 3 m/s2, 7 m/s2, 5 m/s2]

2.9 Spring Force

We know that the more force we apply to a spring, the more it stretches. For a spring that obeysHooke's law, the extension of the spring is proportional to the applied force. Figure-2.104 shows a spring inits equilibrium length. If we stretch it by a distance x from its equilibrium position, it applies a restoringforce F, towards its equilibrium position, which is proportional to x, given by

f = kx (2.59)

Here k is a proportionality constant, known as spring constant. A spring has a tendency of restoringits equilibrium position, thus whether we stretch it or compress, it always opposes the external force in thedirection towards its equilibrium position.

One more point is to be noted that a spring applies the restoring force equally at both of its ends,doesn't matter whether an end is fixed or not.

188 Forces and Newton's Laws of Motion

As shown in figure-2.104, an end of the spring is fixed to wall and other is pulled by applying a force.As the restoring force is directly proportional to the stretch of compression in it, for stretching it by x, weapply a force F on it and for stretching it to double the length to 2x, we have to apply a force double of theprevious value.

Figure 2.104

We take few examples of dynamics, which includes the concept of spring force.

2.9.1 Force Constant of a Spring

In previous section we have discussed about the force exerted by a spring when stretched orcompressed. The spring exerts a force due to its elastic properties. When a spring is stretched or compressedfrom its natural (relaxed) state, its potential energy increases and the work done in stretching or compressingis stored in it in the form of its elastic potential energy. As every system tend to retain its minimum energystate for greater stability, spring tries to restore its natural state, hence applies the restoring force on theexternal agent pulling or pushing the spring. Already we know that the restoring force which is proportionalto the deformation length of the spring as F = kx, where k is the force constant of the spring which dependson the spring shape, material and its elastic properties. If we cut a spring in two equal halves, what will bethe k for each part of the spring ? Or if two springs of force constants k1 and k2 are connected in series,what will be the equivalent force constant of the spring.

For answering above questions, we must know how k depends on length andshape of spring. First we discuss different configurations of spring combinations.

Springs in Series : Look at figure-2.105. Two light springs of force constants k1 andk2 are connected in series and a mass m is hanging from them. Other end of the springis rigidly connected to the ceiling.

We know that the stretch in a spring is proportional to the tension in it. As springsare light, in both the springs, tension remains same as mg. Thus if x1 and x2 are thestretch in the two springs, we have tension in them.

k1x1 = k2x2 = mg (2.60)

Forces and Newton's Laws of Motion 189

If keq is considered as the equivalent force constant of the combined spring, we have

keq (x1 + x2) = mg (2.61)

Substituting the value of x1 and x2 from equation-(2.60) to equation-(2.61), we get

1keq

= 11k

+ 12k

(2.62)

Equation-(2.62) can be generalized for more than two springs connected in series as

1keq

= 11k

+ 12k

+ 13k

+ ............ (2.63)

Where k1, k2, k3, .... are the spring force constants of the respected springs.

Springs in Parallel : As shown in figure-2.106, if two springs are connected in paralleland a mass m is hanging from the combination than both the springs are stretched byequal amounts say x and for equilibrium of mass m we have

k1x + k2x = mg (2.64)

If keq be the equivalent force constant of the combination, we have

keqx = mg (2.65)

From equations-(2.64) and (2.65), we have

keq = k1 + k2 (2.66)

Above equation can be generalized for more than two springs as

keq = k1 + k2 + k3 + ................ (2.67)

Parts of a Spring : If a spring of force constant k of length l is cut in two parts say of l1 and l2, let usassume that new force constants are k1 and k2 for the two parts. If we connect these two parts in series, theequivalent force constant must be initial k. Thus we have

1keq

= 1

1k +

1

2k (2.68)

According to the molecular properties of a spring, the force constant of a part of the spring is inverselyproportional to its length, which gives us

k 1 = cl1

and k2 = cl2

(2.69)

190 Forces and Newton's Laws of Motion

Where c is a positive constant. Substituting the above values of new force constants k1 and k2 inequation-(2.68), we get

1keq

= lc1 +

lc2

or c = kleq

Using value of c in equation-(2.69), we have

k 1 = k lleq

1and k2 =

k lleq

2 (2.70)

Example 2.36

Figure-2.107 shows a block of mass m attached to a spring offorce constant k and connected to ground by two string ofequal lengths making an angle 90 with each other. In relaxedstate natural length of the spring is l. In the situation shownin figure, find the tensions in the two strings.

Solution

As the natural length of spring is l, and in the situation shown in figure-2.108, its length is 3l/2. Thusthe spring is stretched by a distance l/2 hence it exerts a restoring force on block k(l/2) upward as shownin figure-2.108, which shows also the tensions acting on the block along the directions of the strings.As the block is in equilibrium, we can balance all the forces acting on it along horizontal and verticaldirections.

Figure 2.108

Along horizontal direction T1 sin 30 = T2 sin 60

or T1 = 3 T2 (2.71)

Along vertical direction

k

2l = mg + T1 cos30 + T2 cos 60