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A plank of mass '1kg' is placed on the top of a cylinder of mass '2kg' a's shown in the figure. What force should be applied on the plank in the horizontal direction so that there will be no slipping on the contacts. [Given that the coefficient of friction at both the surfaces, $\mu=\frac{1}{2}, g=10 \mathrm{~m} / \mathrm{s}^{2}$ ].

Option: 1
15N

Option: 2
25N

Option: 3
75N

Option: 4
50N

Answers (1)

best_answer

$F-f =m a p \\$

$F-f =m \, 2 p \\$

$2p =\frac{F-f}{m}$

for the cylinder

$f+f^{\prime} =M a_{C \cdot M} \\$

N =M g

for no slipping between the plank and the cylinder.

$a_{p}=a_{c.m}+r \alpha \\$

$a_{p}=2 a_{c \cdot M} \\$

$\frac{f-f}{m}=2\left[\frac{f+f^{\prime}}{M}\right]$

$\because m=1 \mathrm{~kg}, \quad M=2 \mathrm{~kg} \\$

$\frac{F-f}{1}=\frac{\not 2}{\not 2}\left[f+f^{\prime}\right] \\$

$F=2 f+f^{\prime}$

$\because f={\mu mg}$

$f^{\prime}=\mu(m+M) g \\$

$F =2 \mu m g+\mu(m+M) g \\$

$F =2 \mu g\left[m+\frac{m}{2}+\frac{M}{2}\right] \\$

$F =2 \mu g\left[\frac{3 m}{2}+\frac{M}{2}\right]=\mu g(3 m+M) \\$

$F =\frac{1}{2} \times 10 \times 5=25 \mathrm{~N}$

Posted by

Devendra Khairwa

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