Put a uniform meter scale horizontally on your extended index fingers with the left one at 0.00 cm a

Q: Why don't both fingers slip at the same time?

Because the normal forces on the two fingers are different! The finger closer to the center of mass supports more weight, meaning it has a higher maximum static friction. The finger further away will always slip first.

Q: Why does the slipping finger eventually stop?

As the slipping finger moves closer to the center, it supports more and more of the scale's weight. This increases the kinetic friction acting on it. Eventually, this kinetic friction becomes large enough to overcome the maximum static friction of the *other* finger, causing the roles to switch!

Q: How do we know the direction of the friction forces?

Friction opposes relative motion. Since your fingers are moving inward relative to the scale, the friction force exerted *by* the fingers *on* the scale is directed inward, towards the center.

Q: Why do we use torque about the center of mass?

Taking torque about the center of mass eliminates the torque due to gravity ($Mg$), leaving a simple balance between the torques produced by the two normal forces: $N_1 x_L = N_2 x_R$.

Q: What would happen if the coefficients of static and kinetic friction were equal?

If $\mu_s = \mu_k$, the motion would be continuous and jerky, but the math would imply that as soon as one finger reaches the symmetric distance of the other, they would both slide together. The difference between $\mu_s$ and $\mu_k$ is what causes the distinct alternating 'stop-and-go' phases!

Question

Put a uniform meter scale horizontally on your extended index fingers with the left one at 0.00 cm and the right one at 90.00 cm. When you attempt to move both the fingers slowly towards the center, initially only the left finger slips with respect to the scale and the right finger does not. After some distance, the left finger stops and the right one starts slipping. Then the right finger stops at a distance $x_R$ from the center (50.00 cm) of the scale and the left one starts slipping again. This happens because of the difference in the frictional forces on the two fingers. If the coefficients of static and dynamic friction between the fingers and the scale are 0.40 and 0.32, respectively, the value of $x_R$ (in cm) is ______.

Sigma Percentile · Accepted Answer

### The Magic of the Dancing Fingers

Have you ever tried balancing a long stick or a broom on your two index fingers and slowly moving them together? If you haven't, you should! You'll notice something almost magical: your fingers don't slide smoothly together. Instead, they take turns. One finger slides while the other stays perfectly still, and then, as if by some invisible command, they switch roles!

This isn't magic; it's a beautiful symphony of **torque**, **static friction**, and **kinetic friction**. Let's break down exactly why this happens and how we can mathematically predict the exact point where the fingers switch roles.

### Analyzing the Setup

Imagine our uniform meter scale resting horizontally on two fingers. Gravity pulls the center of mass down with a force $Mg$. To keep the scale from falling, your fingers push up with normal forces $N_1$ (left finger) and $N_2$ (right finger).

For the scale to remain perfectly balanced, the net torque about the center of mass must be zero. This gives us our master equation:
$$N_1 x_L = N_2 x_R$$
where $x_L$ and $x_R$ are the distances of the left and right fingers from the center, respectively.

Notice what this equation tells us: the finger closer to the center must exert a **larger normal force** to maintain balance.

### Phase 1: The Left Finger Slips

Initially, the left finger is at $0.00 	ext{ cm}$ ($x_L = 50 	ext{ cm}$ from the center) and the right finger is at $90.00 	ext{ cm}$ ($x_R = 40 	ext{ cm}$ from the center). Because the right finger is closer to the center, $N_2 > N_1$.

Friction is what allows your fingers to slide. The maximum static friction a finger can provide is $\mu_s N$. Since $N_2 > N_1$, the right finger has a much stronger "grip" on the scale. Therefore, when you push inward, the left finger's grip breaks first. It begins to slide, experiencing **kinetic friction** ($f_1 = \mu_k N_1$), while the right finger remains locked in place by **static friction** ($f_2 \le \mu_s N_2$).

To keep the scale from accelerating horizontally, the friction forces must balance:
$$f_1 = f_2$$

### The Switching Point

As the left finger slides closer to the center, its distance $x_L$ decreases. According to our torque equation, this means $N_1$ must increase and $N_2$ must decrease.

Because $N_1$ is increasing, the kinetic friction $f_1 = \mu_k N_1$ is also growing. The right finger has to provide more and more static friction to match it. Eventually, the required friction hits the right finger's absolute limit: $f_2 = \mu_s N_2$.

At this exact moment, the right finger is on the verge of slipping! We can equate the forces:
$$\mu_k N_1 = \mu_s N_2$$

Substituting the given coefficients ($\mu_k = 0.32$ and $\mu_s = 0.40$):
$$0.32 N_1 = 0.40 N_2 \implies 4 N_1 = 5 N_2$$

Now, where is the left finger when this happens? We return to our torque equation:
$$N_1 x_L = N_2 (40)$$
Substituting $N_2 = \frac{4}{5} N_1$:
$$N_1 x_L = \left(\frac{4}{5} N_1ight) (40)$$
$$x_L = 32 	ext{ cm}$$

The left finger stops exactly $32 	ext{ cm}$ from the center!

### Phase 2: The Right Finger Takes Over

Now the roles are reversed. The left finger is locked in place at $32 	ext{ cm}$, and the right finger begins to slide inward from $40 	ext{ cm}$. The right finger experiences kinetic friction ($f_2 = \mu_k N_2$), and the left finger provides static friction ($f_1 \le \mu_s N_1$).

As the right finger moves closer, $N_2$ increases and $N_1$ decreases. The right finger will slide until the kinetic friction it generates matches the maximum static friction of the left finger:
$$\mu_s N_1 = \mu_k N_2$$

Substituting the coefficients again:
$$0.40 N_1 = 0.32 N_2 \implies 5 N_1 = 4 N_2$$

### Final Calculation

To find the final position of the right finger ($x_R$), we use the torque equation one last time. We know the left finger is at $x_L = 32 	ext{ cm}$:
$$N_1 (32) = N_2 x_R$$

Substituting $N_1 = \frac{4}{5} N_2$:
$$\left(\frac{4}{5} N_2ight) (32) = N_2 x_R$$
$$x_R = \frac{128}{5} = \mathbf{25.60 	ext{ cm}}$$

And there we have it! The right finger stops at exactly $25.60 	ext{ cm}$ from the center, and the left finger will begin to slide once again. This beautiful alternating dance continues until both fingers meet perfectly at the center of mass.

Enter Numerical Value:

Visualized Solution

$Free Body Diagram$

$Phase 1: Left Finger Slips$

$Right Finger About to Slip$

$Position of Left Finger (x_{L})$

$Phase 2: Right Finger Slips$

$Left Finger About to Slip Again$

$Final Position of Right Finger (x_{R})$

The Sigma Insight: Torque and Angular Momentum

The Magic of the Dancing Fingers

Analyzing the Setup

Phase 1

The Switching Point

Phase 2

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Free Body Diagram

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Phase 1: Left Finger Slips

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Right Finger About to Slip

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Position of Left Finger (xL​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Phase 2: Right Finger Slips

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Left Finger About to Slip Again

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Position of Right Finger (xR​)