Question:
If the tension in the string in figure is $16 \mathrm{~N}$ and the acceleration of each block is $0.5 \mathrm{~m} / \mathrm{s}^{2}$, find the friction coefficients at the two contacts with the blocks.
Solution:
For $2 \mathrm{Kg}$ block
$T-\mu_{1} N_{1}=2 \times a$
$16-\mu_{1}(20)=2 \times(0.5)$
$15=20 \mu_{1}$
$\mu_{1}=0.75$
For $4 \mathrm{Ka}$ block
$40 \sin 30-T-\mu_{2} N_{2}=4 a$$40 \sin 30-T-\mu_{2} N_{2}=4 a$
$40 \times\left(\frac{1}{2}\right)-T-\mu_{2}(40 \cos 30)=4 a$
$20-16-\mu_{2} \times 40 \times \frac{\sqrt{3}}{2}=4 \times 0.5$
$\mu_{2}=0.06$
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