I did a little experiment with balancing an orange on a spoon. And i realized i don't really understand why i feel a stronger force under both fingers, the further my forefinger is from the orange (As shown in the first photos with the red arrows)

And why i feel a weaker force under both fingers the closer my forefinger is to the orange. (As shown in the second photo with the cyan arrows)

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2 Answers 2


We'll assume that your forefinger serves as the fulcrum or pivot point in what follows.

For the first case: The orange sits at the end of a lever arm. its weight times the length of the lever arm produces a torque which wants to twist the spoon counterclockwise around your forefinger.

To prevent this from occurring, you press your thumb down in contact with the end of the spoon handle to produce a counter-torque: the press-down force times the lever arm (which is the distance between the forefinger and the thumb- which is quite short in the first case) needs to cancel the torque produced by the orange. A short lever arm means high force, and you feel that in your fingers.

Second case: Here, the fulcrum (your forefinger) is closer to the orange, which gives the orange a shorter lever arm. the weight of the orange at the end of the lever arm hence produces less torque, and therefore your fingers can cancel the torque more easily. In addition, your thumb and forefinger are further apart, which lengthens the lever arm for your thumb so it produces more torque for the same force. both of these effects reduce the forces required to cancel the torque from the orange, and you feel the reduction in effort required.

You can model this system as a teeter-totter, with the orange on one end and your thumb on the other, with your forefinger as the pivot. Sliding the pivot towards the orange and away from your thumb simultaneously shortens the orange's lever arm while lengthening your thumb's lever arm. This lets your thumb balance the orange with less downforce.

  • $\begingroup$ But where does this "extra force" come from ? , i mean how can it be that the same weight of the orange "produces" a stronger force, in different positions ? $\endgroup$ Commented Feb 1, 2018 at 5:23
  • $\begingroup$ @physicsnewbie The actual net force you apply to the spoon is the same in both cases. When you are holding it the first way, you are applying a large force downwards and a slightly larger force upwards with the other finger. Basically your fingers are fighting each other. $\endgroup$
    – Chris
    Commented Feb 1, 2018 at 5:29
  • $\begingroup$ you have to think about torques here to get the problem: not force by itself, but force times length of lever arm. $\endgroup$ Commented Feb 1, 2018 at 5:34
  • $\begingroup$ @Chris if you try it now with any object placed on the spoon in the position shown in the first photo, you will feel there is a distinctively larger force working on both fingers which does not feel like the net force of the fingers in the position of the second photo. $\endgroup$ Commented Feb 1, 2018 at 5:37
  • 1
    $\begingroup$ @physicsnewbie same logic as climbing a mountain. If you go straight up you need more force than going around a curved slope, simply because you need to do the same work (difference in potential energy) over a smaller distance. When applied to this spoon case, the important distance is the distance your finger moves to lift the orange. This just happens to be proportional to the lever length if you are pivoting, hence we talk about torque. $\endgroup$ Commented Feb 1, 2018 at 6:23

Three forces are acting in this scenario. Two are downwards: the gravitational force on the orange and the force applied by your thumb. The third is upwards: the force applied by your forefinger. Since the scenario is static, the sum of all three forces is zero (counting for example the upward direction as positive and downward as negative).

$$ F_{orange} + F_{forefinger} + F_{thumb} = 0 $$

or equivalently:

$$ F_{thumb} = - (F_{orange} + F_{forefinger}) $$ The torque also needs to be zero, otherwise the spoon will start to rotate even when the sum of the forces is zero. Torque is defined as the distance to some chosen reference point times the force perpendicular to the line that connects to that reference point. The reference point can be chosen freely, so let's pick the orange as the reference point. The torque relative to this point is then:

$$ D_{forefinger}F_{forefinger} + D_{thumb}F_{thumb} = 0 $$

where $D_{forefinger}$ is the forefinger's distance to the orange, and $D_{thumb}$ is the thumb's distance to the orange. Since the orange has zero distance to itself, the $F_{orange}$ term has been omitted.

Solve this equation for $F_{thumb}$:

$$ F_{thumb} = - \frac{D_{forefinger}}{D_{thumb}} F_{forefinger} $$

Now we can eliminate $F_{thumb}$ and solve for $F_{forefinger}$:

$$ F_{forefinger} = - \frac{D_{thumb}}{D_{thumb}-D_{forefinger}} F_{orange} $$

Here we can see that the magnitude of $F_{forefinger}$ is equal to $F_{orange}$ when $D_{forefinger}$ is exactly zero, i.e. when the forefinger is placed directly under the orange. It gets larger the closer the forefinger is moved towards the thumb, since that makes the factor $\dfrac{D_{thumb}}{D_{thumb}-D_{forefinger}}$ grow. The magnitude of $F_{thumb}$ also grows since the sum of all three forces must be zero, and $F_{orange}$ is not changing.


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