Early in the morning a milkman in a Bihar village carries two cans on a bamboo pole across his shoulder. Today one can is full and the other half full. He does not rest the pole at its middle. He slides it until the full can is closer to his neck, and then the pole rides level and steady. He has never measured anything. Yet his shoulder has found a precise point.
Replacing two forces by one
Several forces on a rigid body can often be replaced by one force with the same effect, their resultant. A force has three things: a size, a direction and a line of action. Adding the forces as vectors gives the first two. The line of action needs one more demand: the resultant must have the same torque as the original forces, about every point.
Take two parallel forces pointing the same way: \(F_1\) at a point A and \(F_2\) at a point B, a distance \(d\) apart. Their resultant is parallel to them, of size \(R = F_1 + F_2\). Let it act at a distance \(L\) from A. About A, the force \(F_1\) has no torque, and \(F_2\) has torque \(F_2 d\). The resultant must match it: \(RL = F_2 d\), so \[ L = \frac{F_2\,d}{F_1 + F_2} \]
With \(F_2 = 36\,\text{N}\), the resultant is \(48\,\text{N}\) at \(L = 36 \times 0.8/48 = 0.6\,\text{m}\) from A. It is \(0.6\,\text{m}\) from the smaller force and \(0.2\,\text{m}\) from the larger. Those distances are in the ratio \(3 : 1\), and the forces are in the ratio \(1 : 3\). The resultant always divides the gap in the inverse ratio of the forces, like the pivot of a balanced see-saw.
Two antiparallel forces of unequal size also have a single resultant, but it lies outside the gap, beyond the larger force. Take \(30\,\text{N}\) up at one point and \(10\,\text{N}\) down \(0.4\,\text{m}\) away. The resultant is \(20\,\text{N}\) up, \(0.2\,\text{m}\) beyond the \(30\,\text{N}\) force, on the side away from the \(10\,\text{N}\). When the two forces are equal, no single force will do: they form a couple.
Parallel forces of \(20\,\text{N}\) and \(60\,\text{N}\) act in the same direction on a rod, along lines \(1.2\,\text{m}\) apart. How far from the \(20\,\text{N}\) force does their resultant act, in metres?
Three forces meet at one point
For forces that all pass through one point P, the resultant also passes through P. About P each force has zero torque, so the resultant must have zero torque about P too. This gives a useful rule for a body held by just three forces. Suppose the three lines are not parallel. Two of them cross at some point P. Take torques about P. Those two forces have no torque there, so the torque of the third about P must be zero on its own. Its line must pass through P as well.
If a rigid body is in equilibrium under exactly three forces that are not all parallel, their lines of action pass through a single point.
The rule often finds a direction with no algebra. A beam held by a hinge and a string feels its weight, the string's pull and the hinge force. The hinge force must point at the place where the weight line crosses the string. The last lesson of this chapter draws that crossing and uses it.
A uniform ladder rests against a smooth wall on a rough floor. Its weight acts at its middle, and the wall pushes horizontally at its top. Through which point must the total force of the floor, normal force and friction together, pass?
Why the weight acts at one point
Gravity pulls on every particle of a body. A particle of mass \(m_i\) at position \(\vec r_i\) feels a force \(m_i\vec g\). These many small forces are all parallel, so they have one resultant, of size \(Mg\), where \(M\) is the total mass. Its line of action follows from the torques. About the origin, the total torque of all the weights is \[ \vec\tau = \sum_i \vec r_i \times m_i\vec g = \Big(\sum_i m_i\vec r_i\Big)\times\vec g = M\vec r_{\text{cm}} \times \vec g = \vec r_{\text{cm}} \times M\vec g \]
Here \(\vec r_{\text{cm}} = \sum m_i\vec r_i / M\) is the position of the centre of mass, the point that the chapter on Centre of Mass finds for many bodies. So the whole weight acts as if it were one force \(M\vec g\) at the centre of mass. The point where the resultant weight acts is the centre of gravity. This is why every figure in this book draws the weight from one point of the body.
The centre of gravity of a body is the point through which the resultant of the gravitational forces on all its particles acts. It stays the same point of the body whatever way the body is turned. In a uniform gravitational field it is the centre of mass.
One step of the derivation took \(\vec g\) out of the sum. That is allowed only if \(\vec g\) is the same at every particle, in a uniform gravitational field. For a body the size of a building, the field is uniform to about one part in a hundred thousand, so the two centres coincide. For a very tall body, the lower parts are pulled a little harder, and the centre of gravity lies slightly below the centre of mass.
A log \(3\,\text{m}\) long lies on the ground. To lift its thin end just off the ground takes an upward force of \(200\,\text{N}\). To lift its thick end takes \(400\,\text{N}\). Find the weight of the log and the position of its centre of gravity.
Let the weight \(W\) act at a distance \(x\) from the thick end, so \(3 - x\) from the thin end. While one end is lifted, the other end rests on the ground and acts as a pivot.
Lifting the thin end, torques about the thick end: \(200 \times 3 = Wx\). Lifting the thick end, torques about the thin end: \(400 \times 3 = W(3 - x)\).
Add the two equations: \(600 + 1200 = 3W\), so \(W = 600\,\text{N}\). Then \(x = 600/600 = 1\,\text{m}\).
The log weighs \(600\,\text{N}\), and its centre of gravity is \(1\,\text{m}\) from the thick end. The thick end is harder to lift because the weight acts nearer to it.
A pole \(4\,\text{m}\) long lies on the ground. It takes \(150\,\text{N}\) to lift one end, A, just off the ground, and \(250\,\text{N}\) to lift the other end, B. How far from end B is its centre of gravity, in metres?
Finding the centre of gravity by hanging
A flat plate hung freely from a peg feels two forces: the push of the peg and its weight at the centre of gravity. Two forces balance only if they lie along one line. So the plate turns until its centre of gravity lies straight below the peg. A plumb line hung from the same peg then passes through the centre of gravity, and it can be marked on the plate.
Hang the plate from a second hole, and mark a second line. The two lines cross at the centre of gravity. In Fig 4.1 the crossing G lies outside the material, in the empty corner of the L. A centre of gravity need not lie inside a body. A ring, a horseshoe and a boomerang all have theirs in empty space.
An irregular piece of plywood is hung from a nail through a hole near its edge and allowed to settle. What is certain about its centre of gravity?
Now return to the milkman's pole. Say the full can weighs \(200\,\text{N}\) and the half-full can \(100\,\text{N}\), at the ends of a light pole \(1.5\,\text{m}\) long. The two weights are parallel, so their resultant is \(300\,\text{N}\). It acts where \(300\,L = 100 \times 1.5\), that is \(0.5\,\text{m}\) from the full can. The shoulder must sit exactly there, and it pushes up with \(300\,\text{N}\).
During the Kanwar Yatra, pilgrims carry Ganga water in two pots slung from the ends of a decorated bamboo pole, the kanwar. The pole must stay level on the shoulder for hours. If the pots are equal, the shoulder sits at the middle. If one pot holds more, the shoulder moves towards it, to the point that divides the pole in the inverse ratio of the weights.
- Placing the resultant of parallel forces nearer the smaller force.The resultant lies nearer the larger force; the distances are in the inverse ratio of the forces.
- Splitting the gap in the ratio of the forces instead of the inverse ratio.Use \(RL = F_2 d\) about the line of \(F_1\), and let the arithmetic place it.
- Assuming the centre of gravity must lie inside the body.A ring, an L-shaped plate or a horseshoe has its centre of gravity in empty space.
- Using the three-force rule with parallel forces.Parallel lines never meet. For parallel forces, the balance of torques decides the positions instead.