# Why is moment dependent on the distance from the point of rotation to the force?

The formula for moment is:

$$M = Fd$$

Where F is the force applied on the object and d is the perpendicular distance from the point of rotation to the line of action of the force.

Why? Intuitively, it makes sense that moment is dependent on force since the force "increases the intensity". But why distance? Why does the distance from the line of action of the force to the point of intensity affect the moment?

I am NOT looking for a derivation of the above formula from the cross product formula, I am looking for intuition. I understand how when I am turning a wrench, if the wrench is shorter its harder to turn it but I don't understand WHY.

Thanks.

## 3 Answers

Why?

Because a moment is a manifestation of a force at a distance, the same way the a velocity is a manifestation of a rotation at a distance. Given two points A and B you know that $$\vec{M}_A = \vec{r}_{AB} \times \vec{F}_B \\ \vec{v}_A = \vec{r}_{AB} \times \vec{\omega}_B$$

The force at B causes a torque at A, simarly to how a rotation at B causes velocity at A.

So Why is that?

Both forces/torques and velocities/rotations are 3D screws that contain the following properties. a) A line of direction, b) a magnitude, c) a pitch. Forget about the b) and c) for now and focus on the line.

How do you describe a line in 3D. A line has 4 degrees of freedom, and it is usually represented using 6 components with something called Pluecker coordinates. There involve two vectors, each with 3 components. The first vector, I call $\vec{F}$ gives the direction of line line, but its magnitude is not important. So two degrees of freedom are used from the vector. The second vector, I call $\vec{M}$ gives the moment of the line about the origin and it is used to describe the closest point of the line to the origin. It too uses two degrees of freedom because the location along the line is unimportant. It represents either a) The moment of a force along the line, or b) the speed of a rotating body about the line. The location of the line is given by

$$\vec{r} = \frac{\vec{F} \times \vec{M}}{\vec{F} \cdot \vec{F}} = - \frac{\vec{M} \times \vec{F}}{\vec{F} \cdot \vec{F}}$$ depending on which you like best.

Similarly for motions

$$\vec{r} = \frac{\vec{\omega} \times \vec{v}}{\vec{\omega} \cdot \vec{\omega}} = - \frac{\vec{v} \times \vec{\omega}}{\vec{\omega} \cdot \vec{\omega}}$$.

So the moment is a manifestation of a line at a distance.

• @dfg Thank you for accepting the answer. The information provided is at the graduate level for robotics and it is not easy to directly comprehend. I hope this answer has been of some use to you. – John Alexiou Oct 12 '13 at 22:32
• Thank you for the answer! The answer has definitely been a lot of use to me. – dfg Oct 13 '13 at 14:12

I understand how when I am turning a wrench, if the wrench is shorter its harder to turn it but I don't understand WHY.

Suppose a bolt can be unscrewed with one turn, and the process consumes $E$ Joules. Then since $w=F d$, we have $$E=2\pi rF.$$ Thus $$F=\frac{E}{2\pi r}.$$ That's why it's harder to unscrew a bolt using a short wrench. You need to push harder. Does this answer your question?

The reason why torque (rotational force) depends on the distance $d$ from the pivot of rotation (i.e. why torque is a moment) is the following:

Torque is defined as change in angular momentum; if mass is constant that means change in angular velocity.

To achieve a change in angular velocity using a tangential force $F$, we need to travel a greater distance when we are farther away from the center. Or in other words, force $F$ only changes linear velocity; achieving a change in angular velocity requires more the farther out we are.