Why the Door Handle Is as Far From the Hinge as Possible

Try pushing a door open with your hand placed right beside the hinge. It takes a genuinely large effort. Move your hand to the outer edge and the same door swings open with a fingertip.

The force you can apply has not changed. What changed is its distance from the hinge — and the turning effect depends on both.

So a force can be described completely by its size and direction and still not tell you whether a door will open. This page covers the second part of the ICSE Class 8 Physics chapter on force and pressure: the turning effect of a force, the moment that measures it, and the principle that balances two moments against each other.

Because the same moment can be produced by a smaller force acting at a greater distance.

Suppose a tight nut needs a moment of $5\text{N}\text{m}$ to turn. With a short spanner of arm $0.25\text{m}$:

$$F=\frac{\text{moment}}{d}=\frac{5}{0.25}=20\text{N}$$

Swap it for a spanner of arm $0.50\text{m}$:

$$F=\frac{5}{0.50}=10\text{N}$$

Half the effort for the same result. Doubling the arm halves the force needed, because their product must stay at $5\text{N}\text{m}$.

This explains a long list of everyday designs.

- A door handle on the edge farthest from the hinge. For a door needing a moment of $12\text{N}\text{m}$, a handle $0.8\text{m}$ from the hinge needs $12÷0.8=15\text{N}$, while a push $0.1\text{m}$ from the hinge would need $12÷0.1=120\text{N}$ — eight times as much.
- A tap or a screw cap made wide rather than narrow, so fingers act at a larger radius.
- A steering wheel of large diameter, and a spanner fitted with a pipe when a nut will not budge.
- Long-handled scissors and pliers for cutting stiff material near the joint, where the distance from the pivot to the material is small and the distance to your hand is large.
- A heavy door given a longer handle than a light one.

And the reverse case. A bicycle brake lever and a nutcracker work the same way: your hand acts far from the pivot and the load sits close to it, so a modest grip produces a large squeeze on the brake cable or the shell.

When a body is balanced about a pivot, the total clockwise moment equals the total anticlockwise moment:

$$F_1\times d_1=F_2\times d_2$$

This is the principle of moments, and it is the condition for rotational balance — the counterpart to the balanced forces of the previous part.

Worked example — a see-saw. A child of weight $300\text{N}$ sits $1.2\text{m}$ from the pivot. Where must a $400\text{N}$ child sit to balance?

Anticlockwise moment from the first child:

$$300\times 1.2=360\text{N}\text{m}$$

For balance, the clockwise moment must also be $360\text{N}\text{m}$:

$$400\times d=360,d=\frac{360}{400}=0.9\text{m}$$

So the heavier child sits closer to the pivot — $0.9\text{m}$ against $1.2\text{m}$.

Checking: $300\times 1.2=360$ and $400\times 0.9=360$. Equal, so the see-saw balances.

Reading the result. The heavier child always sits nearer. Weight and distance are in inverse proportion here: because the product is fixed, raising one must lower the other. Two children of equal weight balance at equal distances, which is why an evenly matched pair sits at the ends.

Worked example — finding an unknown weight. A uniform rod is pivoted at its centre. A weight of $50\text{N}$ hangs $0.6\text{m}$ to the left, and an unknown weight $W$ hangs $0.4\text{m}$ to the right:

$$50\times 0.6=W\times 0.4$$

$$30=0.4W,W=75\text{N}$$

The shorter arm carries the larger weight, as it must.

What happens when they are not equal. If the clockwise moment exceeds the anticlockwise one, the body rotates clockwise. A see-saw with the heavier child too far out simply goes down on that side — the moments are unbalanced and rotation follows, just as an unbalanced force produces motion.