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4-1.Newton's Laws of Motion
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Two wooden blocks are moving on a smooth horizontal surface such that the mass $m$ remains stationary with respect to block of mass $M$ as shown in the figure. The magnitude of force $P$ is
A$(M + m)g\, tan\,\beta $
B$g\,tan\, \beta $
C$mg\,cos \,\beta $
D$(M + m)g \,coses\,\beta $
Solution
The different forces acting on mass $\mathrm{m}$ are shown in the adjoining figure.
Acceleration of the system
$=\frac{P}{M+m}$
$\therefore$ Force on mass $\mathrm{m}=\frac{\mathrm{Pm}}{\mathrm{M}+\mathrm{m}}$
Let the reaction of $\mathrm{m}$ on $\mathrm{M}$
be $\mathrm{f}$. Then $\mathrm{f}=\frac{\mathrm{Pm}}{\mathrm{M}+\mathrm{m}}$
According to figure, $\mathrm{m}$ will be stationary when
${\text { f } \cos \beta=\operatorname{mg} \sin \beta}$
${\frac{\operatorname{Pm}}{M+m} \cos \beta=\operatorname{mg} \sin \beta}$
${P=(M+m) g \tan \beta}$
Acceleration of the system
$=\frac{P}{M+m}$
$\therefore$ Force on mass $\mathrm{m}=\frac{\mathrm{Pm}}{\mathrm{M}+\mathrm{m}}$
Let the reaction of $\mathrm{m}$ on $\mathrm{M}$
be $\mathrm{f}$. Then $\mathrm{f}=\frac{\mathrm{Pm}}{\mathrm{M}+\mathrm{m}}$
According to figure, $\mathrm{m}$ will be stationary when
${\text { f } \cos \beta=\operatorname{mg} \sin \beta}$
${\frac{\operatorname{Pm}}{M+m} \cos \beta=\operatorname{mg} \sin \beta}$
${P=(M+m) g \tan \beta}$
Standard 11
Physics
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