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JEE Advanced 2020 · Paper 1, Physics question 13

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 xRx_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 xRx_R (in cm) is ______.

Answer and worked solution

Official answer: 25.6

Measure positions from the centre of mass (the 50 cm mark). Torque balance on the scale gives NLxL=NRxRN_Lx_L=N_Rx_R, so the finger nearer the centre carries more load.

First stage. The left finger starts 50 cm away and the right 40 cm away, so NL<NRN_L<N_R and the left finger slips. It slides in, with kinetic friction μkNL\mu_kN_L balanced by static friction on the right, until
μkNL=μsNR  ⟹  xRxL=NLNR=0.400.32=1.25  ⟹  xL=401.25=32 cm.\mu_kN_L=\mu_sN_R\implies\frac{x_R}{x_L}=\frac{N_L}{N_R}=\frac{0.40}{0.32}=1.25\implies x_L=\frac{40}{1.25}=32\text{ cm}.

Second stage. Now the right finger slips while the left (at 32 cm) holds. It stops when
μkNR=μsNL  ⟹  xLxR=1.25  ⟹  xR=321.25=25.6 cm.\mu_kN_R=\mu_sN_L\implies\frac{x_L}{x_R}=1.25\implies x_R=\frac{32}{1.25}=25.6\text{ cm}.
Answer: 25.60

Topics:FrictionEquilibrium of rigid bodies
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