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(A) Let $N_1$ and $N_2$ be the normal forces on the left and right fingers respectively. For a uniform meter scale of mass $M$ and length $100 \ cm$,t

Vedclass · Accepted Answer

$25.60$

Vedclass · Answer

(A) Let $N_1$ and $N_2$ be the normal forces on the left and right fingers respectively. For a uniform meter scale of mass $M$ and length $100 \ cm$,the center of mass is at $50 \ cm$. Initially,the fingers are at $0 \ cm$ and $90 \ cm$. The distance of the center of mass from the left finger is $50 \ cm$ and from the right finger is $40 \ cm$.For rotational equilibrium about the center of mass: $N_1(50) = N_2(40) \implies 5N_1 = 4N_2$.Also,$N_1 + N_2 = Mg$. Substituting $N_2 = 1.25N_1$,we get $2.25N_1 = Mg \implies N_1 = \frac{4}{9}Mg$ and $N_2 = \frac{5}{9}Mg$.When the left finger slips,it experiences kinetic friction $f_{k1} = \mu_k N_1$ and the right finger experiences static friction $f_{s2} \le \mu_s N_2$. When the left finger stops and the right starts slipping,the right finger experiences kinetic friction $f_{k2} = \mu_k N_2$ and the left finger experiences static friction $f_{s1} = \mu_s N_1$.At the point where the right finger stops and the left finger starts slipping,the torque about the center of mass is zero: $N_1 x_L = N_2 x_R$.Also,the condition for the transition is $f_{s1} = f_{k2} \implies \mu_s N_1 = \mu_k N_2$.Given $\mu_s = 0.40$ and $\mu_k = 0.32$,we have $0.40 N_1 = 0.32 N_2 \implies N_1 = 0.8 N_2 = \frac{4}{5} N_2$.Substituting this into the torque equation: $(\frac{4}{5} N_2) x_L = N_2 x_R \implies x_R = 0.8 x_L$.From the previous stage where the left finger stopped,$N_1 x_L = N_2(40)$ with $4N_1 = 5N_2$ (i.e.,$N_1 = 1.25 N_2$),we found $x_L = 32 \ cm$.Thus,$x_R = 0.8 	imes 32 = 25.6 \ cm$.

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