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 ______.

Answer 1

let N₁ and N₂ are the normal reactions acting on the scale at both ends as shown in figure. f₁ and f₂ are frictional forces and xL is the distance of left finger when right finger starts moving.

solution : initially at equilibrium,

vertically upward force = vertically downward force

⇒N₁ + N₂ = Mg .....(1)

now torque about centre = 0

⇒N₁ × 50 + (-N₂) × 40 = 0

⇒5N₁ = 4N₂ ......(2)

from equations (1) and (2) we get,

N₁ = 4Mg/9 and N₂ = 5Mg/9

for 1st ,

static friction force , f₁k = = μ_k N₁ = 0.32 N₁

similarly f₂k = μ_k N₂ = 0.32 N₂

dynamic friction force, f₁L = μ_s N₁ = 0.4 N₁

similarly, f₂L = μ_s N₂ = 0.4 N₂

now torque about centre = 0,

N₁xL + (- N₂)(40) = 0......(3)

⇒f₁k = f₂L

⇒0.32N₁ = 0.4 N₂

⇒4N₁ = 5N₂

so, N₁xL = 4N₁/5 (40)

⇒xL = 32 cm

Now xR = distance when right finger stops and left starts moving.

torque about centre = 0

so, N₁xL = N₂xR .......(4)

f₁L = f₂k ⇒0.4 N₁ = 0.32N₂

5N₁ = 4N₂

from equation (4),

N₁ × 32 = 5N₁/4 × xR

⇒32 × 4/5 = xR

⇒xR = 25.6 cm

Therefore the value of xR is 25.6 cm