8.14 Spring (application)  (Page 2/2)

$$\Rightarrow \left|a\right|=\frac{\left(k_1+k_2\right)x}{m}$$

Spring and pulley

Problem 3 : A spring of spring constant 200 N/m is attached to a “block – pulley” system as shown in the figure. Find the extension in the spring.

Solution : We have studied that there is two phases of extension/compression of a spring. One is a brief period in which spring dynamically changes its length depending on the net axial force on it. This means that forces at the two ends of the spring are not same. Quickly, however, spring acquires the stabilized extension in the second phase, usually denoted by symbol “ $x_0$ ”. In this stabilized condition, the forces at the two ends of an ideal (without mass) spring are equal.

This problem deals the situation in which spring has acquired the stabilized extension. Thus forces across the spring are equal at two ends and are also equal to the tensions in the attached string at those ends. In the force analysis, we shall treat spring just like string with the exception that the spring force equals tension in the string. Hence,

$$T=k{x}_0$$

As the masses of the blocks are equal, the net pulling force on the blocks in the downward direction is zero.

$$F=5g-5g=0$$

It means that blocks have zero acceleration. Now, free body diagram (as shown on the right hand side of the figure above) of any of the given blocks yields :

$$\Rightarrow T-5g=mXa=0$$

$$\Rightarrow T=5g=k{x}_0$$

$$\Rightarrow x_0=\frac{5g}{k}=\frac{5X10}{200}=0.25m$$

Extension in spring

Problem 4 : A spring of spring constant "k", has extended lengths " $x_1$ " and " $x_2$ " corresponding to spring forces 4 N and 5 N. Find the spring length when spring force is 9 N.

Solution : Let the natural length of spring be “x”. According to Hooke’s law, the magnitude of spring force,

$$F=k\Delta x$$

where " $\Delta x$ " is the extension. Let the spring length for spring force 9N be " $x3$ ". Then,

$$9=k\left(x_3-x\right)=k{x}_3-kx$$

$$\Rightarrow x_3=\frac{9+kx}{k}$$

We need to find expressions of spring constant and natural length in terms of given values. For the given two extensions, we have :

$$4=k\left(x_1-x\right)$$

$$5=k\left(x_2-x\right)$$

Subtracting first equation from second and solving for “k”,

$$\Rightarrow 1=k\left(x_2-x_1\right)$$

$$k=\frac{1}{\left(x_2-x_1\right)}$$

Substituting for “k” in the first equation,

$$\Rightarrow 4=\frac{1}{\left(x_2-x_1\right)}X\left(x_1-x\right)$$

$$\Rightarrow 4\left(x_2-x_1\right)=\left(x_1-x\right)$$

$$\Rightarrow 4x_2-4x_1=x_1-x$$

$$\Rightarrow x=5x_1-4x_2$$

We can, now, evaluate the required expression of “ $x_3$

$$\Rightarrow x_3=\frac{9+\frac{1}{\left(x_2-x_1\right)}X\left(5x_1-4x_2\right)}{\frac{1}{\left(x_2-x_1\right)}}$$

$$\Rightarrow x_3=9\left(x_2-x_1\right)+\left(5x_1-4x_2\right)$$

$$x_3=5x_2-4x_1$$

Spring force as contact force

Problem 5 : A bead of mass, “m”, is placed on a circular rim of radius “r” and is attached to a fixed point “A” through a spring as shown in the figure. The natural length of spring is equal to the radius “r” of the circular rim and spring constant is “k”. Find the normal force on the circular rim at the moment spring is released from the position “B”. The spring makes 30° with the horizontal in this position.

Solution : The forces on the bead are (i) its weight (ii) spring force and (iii) normal force.

In order to carry force analysis, we need to determine magnitudes and directions of forces. Here, we know the directions of all three forces. Spring force acts along spring length, gravity acts vertically downward and normal force acts radially. In triangle ABC, CA and CB are the radii. Hence, triangle ABC is an isosceles triangle.

$$\angle CAB=\angle CBA={30}^0$$

Let "F" be the spring force. Normal force is equal to sum of the components of spring force and gravity in radial direction. Carrying force analysis in CB direction, we have :

$$N=mg\cos {30}^0+F\cos {30}^0$$

The magnitude of spring force, “F”, is given by :

$$F=kx$$

where “x” is the extension in the spring and is given by :

$$x=AB-r=2r\cos {30}^0-r=\left(\sqrt{3}-1\right)r$$

Substituting in the expression of normal force, we have :

$$N=mgX\frac{\sqrt{3}}{2}+k\left(\sqrt{3}-1\right)r\sqrt{\frac{3}{2}}$$

$$N=\frac{\sqrt{3}}{2}\{mg+k\left(\sqrt{3}-1\right)r\}$$