JEE Challenger
More from Laws of Motion
PhyMedAdv 2020 • Paper 1

Frictional Slip Dynamics of a Meter Scale on Fingers

Place a uniform meter scale horizontally across two index fingers, with the left finger positioned at 0.00 cm0.00\text{ cm} and the right finger at 90.00 cm90.00\text{ cm}. As you slowly slide both fingers toward the center, initially only the left finger slips relative to the scale, while the right finger stays stationary. After covering a certain distance, the left finger halts and the right finger starts to slip. The right finger then comes to rest at a distance xRx_R from the center (50.00 cm50.00\text{ cm}) of the scale, after which the left finger begins slipping again. This sequence is caused by the difference between static and kinetic frictional forces. If the static and dynamic friction coefficients between the fingers and the scale are 0.400.40 and 0.320.32 respectively, find the value of xRx_R (in cm\text{cm}).

Official Numerical Answer25.6

Topics & Concepts

Laws of MotionFriction

Step-by-Step Solution

To find the distance xRx_R of the right finger from the center of the meter scale when it comes to rest, we analyze the conditions for rotational and translational equilibrium along with the friction forces acting at both fingers.

1. Coordinate Setup and Initial Conditions

Let the center of mass of the uniform meter scale be at the origin, i.e., at the 50.00 cm50.00\text{ cm} mark.

  • Initial position of the left finger: 0.00 cm  ⟹  0.00\text{ cm} \implies Distance from the center, dL=50.00 cmd_L = 50.00\text{ cm}.
  • Initial position of the right finger: 90.00 cm  ⟹  90.00\text{ cm} \implies Distance from the center, dR=40.00 cmd_R = 40.00\text{ cm}.

Let NLN_L and NRN_R be the normal forces exerted by the left and right fingers, respectively, and let mm be the mass of the scale.

From vertical equilibrium: NL+NR=mgN_L + N_R = mg

From rotational equilibrium about the center of mass: NLdL=NRdRN_L d_L = N_R d_R

2. Phase 1: Left Finger Slipping, Right Finger Stationary

Initially, dL=50.00 cmd_L = 50.00\text{ cm} and dR=40.00 cmd_R = 40.00\text{ cm}. Since dL>dRd_L > d_R, it follows that NL<NRN_L < N_R. Consequently, the maximum static friction force available at the left finger (μsNL\mu_s N_L) is smaller than that at the right finger (μsNR\mu_s N_R). Therefore, the left finger begins to slip first while the right finger remains stationary at dR=40.00 cmd_R = 40.00\text{ cm}.

While the left finger is slipping:

  • Kinetic friction force at the left finger: fL=μkNLf_L = \mu_k N_L
  • Static friction force at the right finger: fR=fL=μkNLf_R = f_L = \mu_k N_L

As the left finger moves inward, dLd_L decreases, which increases NLN_L and decreases NRN_R. The left finger stops slipping and the right finger begins to slip when the static friction required at the right finger reaches its maximum limit: fR=μsNR  ⟹  μkNL=μsNRf_R = \mu_s N_R \implies \mu_k N_L = \mu_s N_R

Using the rotational equilibrium relation NLdL=NRdRN_L d_L = N_R d_R, we substitute NLNR=dRdL\frac{N_L}{N_R} = \frac{d_R}{d_L}: μk(dRdL)=μs  ⟹  dL=(μkμs)dR\mu_k \left(\frac{d_R}{d_L}\right) = \mu_s \implies d_L = \left(\frac{\mu_k}{\mu_s}\right) d_R

Substituting the given values (μs=0.40\mu_s = 0.40, μk=0.32\mu_k = 0.32, and dR=40.00 cmd_R = 40.00\text{ cm}): dL=0.320.40×40.00 cm=0.80×40.00 cm=32.00 cmd_L = \frac{0.32}{0.40} \times 40.00\text{ cm} = 0.80 \times 40.00\text{ cm} = 32.00\text{ cm}

Thus, the left finger comes to rest at a distance of dL=32.00 cmd_L = 32.00\text{ cm} from the center.

3. Phase 2: Right Finger Slipping, Left Finger Stationary

Now, the left finger remains fixed at dL=32.00 cmd_L = 32.00\text{ cm}, and the right finger slips towards the center (so dRd_R decreases).

While the right finger is slipping:

  • Kinetic friction force at the right finger: fR=μkNRf_R = \mu_k N_R
  • Static friction force at the left finger: fL=fR=μkNRf_L = f_R = \mu_k N_R

The right finger stops slipping when the static friction required at the left finger reaches its maximum limit: fL=μsNL  ⟹  μkNR=μsNLf_L = \mu_s N_L \implies \mu_k N_R = \mu_s N_L

Using the rotational equilibrium relation NLdL=NRdRN_L d_L = N_R d_R, we substitute NRNL=dLdR\frac{N_R}{N_L} = \frac{d_L}{d_R}: μk(dLdR)=μs  ⟹  dR=(μkμs)dL\mu_k \left(\frac{d_L}{d_R}\right) = \mu_s \implies d_R = \left(\frac{\mu_k}{\mu_s}\right) d_L

Substituting dL=32.00 cmd_L = 32.00\text{ cm}: xR=dR=0.320.40×32.00 cm=0.80×32.00 cm=25.60 cmx_R = d_R = \frac{0.32}{0.40} \times 32.00\text{ cm} = 0.80 \times 32.00\text{ cm} = 25.60\text{ cm}

Final Answer

The distance xRx_R of the right finger from the center of the scale is 25.60 cm25.60\text{ cm} (or 25.6 cm25.6\text{ cm}).

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