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Block on a Block Problems

For the figur...

Question

For the figure shown below, masses $m_{1}, m_{2}$ and $M$ are $20 kg, 5 kg$ and $50 kg$ respectively. The coefficient of friction between $M$ and ground is zero. while the coefficient of friction between $M$ and $m_{1}$ , and that between $m_{2}$ and ground is $0.3$ . The pulleys and strings are massless. The string is perfectly horizontal between $P_{1}$ and $m_{1}$ , and also between $P_{2}$ and $m_{2}$ . The string is perfectly vertical between $P_{1}$ and $P_{2}$ . An external horizontal force $F$ is applied on the mass $M$ . Take $g = 10 {m/s}^{2}$ . Let the magnitude of force of friction between $M$ and $m_{1}$ be $f_{1}$ and between $m_{2}$ and ground is $f_{2}$ . For a particular $F$ it is found that $f_{1} = 2 f_{2}$ . Choose the correct option(s).

Force acting on block of mass

M

60 N

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Tension in the string is

18 N

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Friction force

f_{1}

and

f_{2}

are

30 N

and

15 N

respectively.

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The acceleration of the block of mass

M

0.3

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Solution

The correct options are
A Force acting on block of mass $M$ is $60 N$ .
B Tension in the string is $18 N$ .
C Friction force $f_{1}$ and $f_{2}$ are $30 N$ and $15 N$ respectively.
The FBDs of the system is as shown

From the FBDs we have
For $m_{2}$ :

$N_{2} = 5 g = 50 N$
$f_{2 max} = μ N_{2} = (0.3) (50) = 15 N$
For $m_{1} :$

$N_{1} = 20 g = 200 N$
$f_{1 max} = μ N_{1} = (0.3) (200) = 60 N$

From here, we can say that $f_{2}$ will reach maximum first
Since, $f_{1} = 2 f_{2} \Rightarrow f_{1} = 30 N$ and the friction between $M$ & $m_{1}$ is static in nature.

Case (i): All blocks are at rest.
For $m_{1} :$ $f_{1} - T = 20 (0)$
For $m_{2} :$ $T - f_{2} = 5 (0)$
_________________
$\Rightarrow f_{1} = f_{2}$ , but given $f_{1} = 2 f_{2}$ . Thus,
the system is in motion and $f_{1}$ is static in nature.
Hence, $m_{1}$ & $m_{2}$ will move together with common acceleration $a$ .
For $m_{1} :$ $f_{1} - T = m_{1} a$
For $m_{2} : T - f_{2} = m_{2} a$
$\Rightarrow f_{1} - f_{2} = a (m_{1} + m_{2})$
$\Rightarrow a = \frac{30 - 15}{25} \Rightarrow a = \frac{3}{5} = 0.6 {m/s}^{2}$
and $T = m_{2} a + f_{2} = 5 \times 0.6 + 15 = 18 N$

Now, for $M$ & $m_{1}$ , taking both as a system
$F - T = (M + m_{1}) a$
$\Rightarrow F = 70 \times 0.6 + 18 = 60 N$

Hence, we get
$F = 60 N, T = 18 N, a = 0.6 {m/s}^{2}$ $f_{1} = 30 N, f_{2} = 15 N$

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