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4.2 friction — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysics4.2 friction1 Mark
MCQ
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. . . . . . . 
  • ✓
    $25.60$
  • B
    $25.65$
  • C
    $25.70$
  • D
    $25.75$
✓

Answer

Correct option: A.
$25.60$
a
Initially

$N _1+ N _2= Mg  N _1=\frac{4 Mg }{9}$

$\left(\tau_{ N }=0\right) N _1(50)= N _2(40)  N _2=\frac{5 Mg }{9}$

$5 N _1=4 N _2$

$f_{l_k}=\mu_k N_1 \quad f_{l_L}=\mu_S N_1$

$f_{i_k}=0.32 N _1 \quad f_{l_1}=0.4 N _1$

$\overline{f_{2 k}=0.32 N _2} \quad \overline{f_{2 L }=0.4 N _2}$

Suppose $x _{ L }=$ distance of left finger from centre when right finger starts moring

$\left(\tau_{ n }=0\right)_{\text {about cratre }} \Rightarrow N _1 x _{ L }= N _2(40)$

$f_{ K _1}= f _{ L _2} \quad \Rightarrow 0.32 N _1=0.40 N _2$

$4 N _1=5 N _2$

$N _{ I } x _{ L }=\frac{4 N _1}{5}(40)$

$x _{ L }=32$

Now $x _{ R }=$ distance when right finger stops and left finger starts moring

$\left(\tau_{ n }=0\right)_{\text {atout centre }} \Rightarrow N _1 x _{ L }= N _2\left( x _{ R }\right)$

$f _{ L _1}= f _{ K _2} \Rightarrow 0.4 N _1=0.32 N _2$

$5 N _1=4 N _2$

$\frac{4 N _2}{5}(32)= N _2 x _{ R }$

$x _{ R }=\frac{128}{5}=25.6 cm$

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