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A block $P$ of mass $M_P$ is in contact with another block $Q$ of mass $M_Q$ as shown in the fígure and they are placed on a smooth floor. Force on block Qis
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Verified Answer
The correct answer is:
$\frac{M_Q F}{M_p+M_Q}$
If $a$ be the common acceleration of system of blocks on application of force $F$.
Free body diagram of block $P$ and $Q$ are given as
Hence, for block $P, F-R=M_P a$
For block $Q, R=M_Q a$
Adding Eqs. (i) and (ii), we get
$$
\Rightarrow \quad F=\frac{\left(M_P+M_Q\right) a}{M_P+M_Q}
$$
$\therefore$ Force on block $Q$ is given as
$$
R=M_Q a=M_Q \cdot \frac{F}{M_P+M_Q}=\frac{M_Q F}{M_P+M_Q}
$$
Free body diagram of block $P$ and $Q$ are given as
Hence, for block $P, F-R=M_P a$
For block $Q, R=M_Q a$
Adding Eqs. (i) and (ii), we get
$$
\Rightarrow \quad F=\frac{\left(M_P+M_Q\right) a}{M_P+M_Q}
$$
$\therefore$ Force on block $Q$ is given as
$$
R=M_Q a=M_Q \cdot \frac{F}{M_P+M_Q}=\frac{M_Q F}{M_P+M_Q}
$$
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