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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.

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8 Answers 8

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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.

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    $\begingroup$ @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. $\endgroup$ Oct 12, 2013 at 22:32
  • $\begingroup$ Thank you for the answer! The answer has definitely been a lot of use to me. $\endgroup$
    – dfg
    Oct 13, 2013 at 14:12
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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?

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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.

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  • $\begingroup$ The net torque causes the rate of change of angular momentum. (Torque times time gives the change.) $\endgroup$
    – R.W. Bird
    Sep 13, 2021 at 18:47
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The best definition of torque (or moment) is the work per unit angle of rotation (in Joules per radian) that can be done by a force which is acting in a manner that tends to cause a rotation. This implies that you want the component of the force which is acting along an arc, times the arc length, divided by the angle (in radians). But the arc length s = rθ, and s/θ = r. So you end up multiplying a force component by a radius.

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Intuitively, torque is the amount of "rotational oomph" you can give to something. The principle you mention is the one by how a wrench works: it increases the $r$ from your hand and whatever you're trying to wrench, e.g. a bolt - and thus increases the torque.

Or conversely, if we were given a set of wrenches and told to try turning various bolts with them, we'd notice that long wrenches would turn the bolt more easily than short ones, and if you took a long wrench and used it with your hand close to the bolt, it would not be so easy. Thus if we were then tasked to develop a measure of the bolt-turning power one can exert when wielding a given wrench, one need to take account both of the direct force applied and also the length of the wrench.

The mathematical expression, then, just refines this by describing the precise shape of the dependency. (Also, the sine part, left off in the expression given in the OP, effectively just reflects that it is the perpendicular part of the force that does the torquing. This should make sense, since rotational velocity is always perpendicular to the radial vector. Hence to change rotational velocity, you need a force in that direction.)

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Understand that there is a tradeoff. You can apply a smaller force with a larger wrench, but you have to move it through a larger distance (arc length) to accomplish the same amount of work (force x distance).

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Intuition, I think maybe it is because when you apply force at a further distance, there is more of that object to follow with the movement the force gives. When I shake my pencil while pinching the end of it, it moves a lot more than when I hold it closer to the pencil nib because there is more of the pencil to follow the movement I gave to the pencil when I shake it when it's longer. Imagine an angle of 30 degrees, even though it's narrow, if the lines forming the angle is long, it's pretty wide, I think this is why distance affects moment

Another thing although a small factor might be because if you pinch the pencil closer to the nib, that means some of the pencil is behind your fingers, and the weight of that part of the pencil can make movement decreased cause your fingers have to produce enough force to move the whole pencil for it to swing and move around, the tip of the pencil wont bend for it to swing.

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seems you have asked a curious question i am 14 right kow and solve such questions with my imigination without any complex mathamatically formulas dervation or whaterver. so just imagine a line segment and make two point (pivot) at the starting)and at the outermost side now think of moving that line about that pivot oke{so would see that a force at the corner is moving the line segment in circular motion} now take another line segment make two point (pivot) at the starting)and at the outermost side and 3rd at the center apply force on the outermost poit on the line with the pivot at the center and now you can see the line is moving in a circular dircetion . now apply force a liitle between the point you se less movement and at the center you see no movement . now just again go to first step and move it from the outermost end you would see that the outer side of the line segment has more force than the inner one and if it has more force than it would try to push the pivot backword and hence leading to moving itself forward . and if the force is applied near the pivot then , the inner part has more power and would try to push itself and the pivot furthur hence at the same time it would be pushing it back NOW YOU WOULD SAY THAT THE FORCES ARE BALCENED BUT WHEN THE FORCE IS APPLIED BELOW A CERTAIN POINT IT WOULD GO IN MINUS . MY ADVICE WOULD BE TO UNDERSTAND TORQUE COMPLETE AND THEN APPLY IT TO THIS . AND BE CURIOUS :-)

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    $\begingroup$ First of all, welcome to Physics SE. I suggest you work a little bit on the format of your answer to be more readable and in addition you could also use a spell checker to make sure you don't have so many spelling mistakes. Furthermore, you could possibly provide a sketch to help the OP understand your answer. $\endgroup$
    – ZaellixA
    Sep 13, 2021 at 15:59

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