Question

two blocks of masses 3m and 2m are in contact on a smooth table A force p is applied first horizontally on block of mass 3m and then on a mass 2m .The contact forces between the two blocks in two cases are in ratio​

Answer 1

According to the whole system of masses 3m and 2m, since there's no other external forces, the net force acted upon the system is nothing but P in each case. Hence we can say that,

$\displaystyle\longrightarrow\sf{P=5ma}$

where a is the net acceleration on the system.

Consider the case where the force P is acted upon the block of mass 3m. Let $\displaystyle\sf {R_1}$ be the normal reaction provided by the block of mass 2m to that of mass 3m, which is opposite to P.

Here the net acceleration a is acting rightwards along P. So,

$\displaystyle\longrightarrow\sf{P-R_1=3ma}$

Since $\displaystyle\sf {P=5ma,}$

$\displaystyle\longrightarrow\sf{5ma-R_1=3ma}$

$\displaystyle\longrightarrow\sf{R_1=2ma}$

Consider the case where the force P is acted upon the block of mass 2m. Let $\displaystyle\sf {R_2}$ be the normal reaction provided by the block of mass 3m to that of mass 2m, which is also opposite to P.

Here the net acceleration a is acting leftwards along P. So,

$\displaystyle\longrightarrow\sf{P-R_2=2ma}$

Since $\displaystyle\sf {P=5ma,}$

$\displaystyle\longrightarrow\sf{5ma-R_2=2ma}$

$\displaystyle\longrightarrow\sf{R_2=3ma}$

Hence the ratio of the reactions (contact forces) is,

$\displaystyle\longrightarrow\sf{\dfrac {R_1}{R_2}=\dfrac {2ma}{3ma}}$

$\displaystyle\longrightarrow\sf{\underline {\underline {\dfrac {R_1}{R_2}=\dfrac {2}{3}}}}$

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If two blocks of masses 'x' and 'y' are in contact with each other on a smooth horizontal surface and a constant force F is acted upon two blocks separately, then the ratio of the contact forces in each case is y:x, i.e., the reciprocal of the ratio of their masses.

In this question the masses are in the ratio 3:2, hence the ratio of the contacting forces is 2:3.

Answer 2

Answer:

the answer

is 2:3

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