IF the tension in the string in figure (6-E3) is 16 N and the acceleration of each block is 0.5 $m / s^{2}$ , Find the friction coefficients at the two contacts with the block.

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Solution

From the free body diagram
$μ_{3} R + 1 - 16 = 0$

$\Rightarrow μ_{1} (2 g) + (- 15) = 0 \Rightarrow μ_{3} = \frac{15}{20} = 0.75 A g a i n μ_{2} R_{1} + 4 \times 0.5 = 16 - 4 g sin 30^{| c i r c} = 0 \Rightarrow μ_{2} (20 \sqrt{3}) + 2 + 16 - 20 = 0 \Rightarrow μ_{2} = \frac{2}{20} \sqrt{3} = \frac{1}{17.32} = 0.057 = 0.06$
Hence co-efficients of friction, $μ_{1} = 0.75$ and $μ_{2} = 0.06$

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IF the tension in the string in figure (6-E3) is 16 N and the acceleration of each block is 0.5 m/s2, Find the friction coefficients at the two contacts with the block.

From the free body diagram μ3R+1−16=0 ⇒μ1(2g)+(−15)=0⇒μ3=1520=0.75Againμ2R1+4×0.5=16−4gsin30|circ=0⇒μ2(20√3)+2+16−20=0⇒μ2=220√3=117.32=0.057=0.06 Hence co-efficients of friction, μ1=0.75 and μ2=0.06

IF the tension in the string in figure (6-E3) is 16 N and the acceleration of each block is 0.5 $m / s^{2}$ , Find the friction coefficients at the two contacts with the block.