 |
| Couple |
|
| Two equal and opposite parallel forces acting along different lines on a body constitute a couple. |
| |
| The turning effect of the couple is called its moment and is calculated by the product of either of the forces and the perpendicular distance between them (i.e., between their lines of action). |
| |
| A body acted upon by a couple will rotate in a clockwise direction or anticlockwise direction as shown [Figure (i), (ii) and (iii) shown below]. If the two forces acting on the body have the same line of action, then the moment becomes zero. |
| |
 |
| |
| Two equal, unlike and parallel forces acting along different lines form a couple. |
| |
| Moment of a couple = Force (N) x Perp. Distance (m) between the forces [Figure (iv) shown below]. |
| |
 |
| |
| Moment of the couple = 10 N x 4 m = 40 N m |
| |
|
| Two examples of the turning effect of two equal and opposite forces not acting in the same straight line are the steering wheel and the pedals of a bicycle. In the figure (a) below, the left hand is pulling with force F on the steering wheel while the right hand is pushing with the same force F. The two forces make the wheel turn in an anticlockwise direction. |
| |
 |
| |
| In figure (b) shown above, one pedal is being pushed forward while the other is being pushed back. This rotates the sprocket wheel and the attached chain anticlockwise. Can you think of other everyday examples in which a turning effect or rotation takes place? |
| |
 Examples of Couple |
| |
| In our day-to-day life, we come across many objects which work on the principle of couple. Winding up the spring of a toy car, opening and closing the cap of a bottle, turning of a water tap, cork screws, door key etc. are some of the common examples of couples. |
| |
|
Applications of principle of moment |
| |
| The principle of moment is applied in simple machines like the lever. A beam balance also works on the principle of moments. |
| |
|
A beam balance |
| |
| The physical balance used in the school laboratory is pivoted in the middle with equal arms. The two scale pans of equal weights are hung from the upper edge of wedge shaped supports at either end of the beam. When the beam is raised for weighing, it swings freely about the lower edge of a wedge shaped support in the center. In this position the balance is in equilibrium. |
| |
 |
| |
| Beam balance |
| |
| Because l1 = l2 and m1 = m2, according to the principle of moments, |
| |
| m1 x l1 = m2 x l2 |
| |
| Now if you place a mass of 1 kg in one pan and an unknown mass 'x' on the other pan so that the balance is in equilibrium. |
| |
| then, (m1 + x) l1 = (m2 + 1) l2 |
| |
| As m1 = m2 and l1 = l2 |
| |
 x = 1 kg |
| |
| Let us take another example of two boys carrying a load on a stick between them as shown in figure below. |
| |
 |
| |
| Let us calculate what part of the load each boy carries. |
| |
| To find the upward force exerted by the boy at A, we shall consider the hand of the boy at B as the pivot. |
| |
| Now, the clockwise moment = F1 x 5 m and the anticlockwise moment due to the load 900 N = 900 x 3. |
| |
| If the bar is in equilibrium, then |
| |
| F1 x 5 = 900 x 3 |
| |
F1 = 900 x  = 540 N |
| |
| Hence, the force exerted by the boy = 540 N. |
| |
| But F1 + F2 = 900 N (sum of the downward forces equal to the sum of upward forces). |
| |
| Therefore, F2 = 900 - F1 |
| |
| = 900 - 540 |
| |
| = 360 N |
| |
| The force exerted by the boy at B can also be calculated by using A as a pivot. |
| |
| Therefore, F2 x 5 = 900 x 2 |
| |
or, F1 = 900 x  |
| |
| = 360 N |
| |
