Two-blocks-of-masses-2-kg-and-3-kg-are-kept-on-a-smooth-inclined-plane-A-constant-force-of-magnitude-20-N-is-applied-on-2-kg-block-parallel-to-the-inclined-The-contact-force-between-the-two-blocks-i

Question Number 23399 by Tinkutara last updated on 29/Oct/17

Two blocks of masses 2 kg and 3 kg

are kept on a smooth inclined plane .

A constant force of magnitude 20 N is

applied on 2 kg block parallel to the

inclined . The contact force between

the two blocks is

Commented by Tinkutara last updated on 29/Oct/17

Commented by mrW1 last updated on 29/Oct/17

F_{12} = \frac{M}{m + M} \times F = \frac{3}{2 + 3} \times 20 = 12 N

Commented by Physics lover last updated on 29/Oct/17

C a n y o u p l z e x p l i n y o u r m e t h o d

s i r .

Commented by mrW1 last updated on 30/Oct/17

If a total force F is applied on a total

mass m, the partial force F_{i} which is

applied on a part of mass (m_{i}) is

F_{i} = \frac{m_{i}}{m} \times F

in our example we can also do like this :

acceleration of (m + M) is

a = \frac{F - (m + M) g \times \sin θ}{m + M}

Ma = F_{12} - Mg \times \sin θ

\Rightarrow F_{12} = M (g \times \sin θ + a) = M [g \times \sin θ + \frac{F - (m + M) g \times \sin θ}{m + M}]

= \frac{M}{m + M} \times F

Answered by Physics lover last updated on 29/Oct/17

\Rightarrow a_{t w o b o c k s} = \frac{5 g \cdot S i n 37 ° - 20}{5}

\Rightarrow a = \frac{5 (10) \frac{3}{5} - 20}{5}

\Rightarrow a = 2 m / s^{2} a l o n g t h e i n c l i n e d d o w n w a r d s

f o r 3 k g b l o c k

\Rightarrow m a = m g \cdot S i n 37 ° - F_{c o n t a c t}

\Rightarrow 3 (2) = 3 (10) \frac{3}{5} - F_{c o n t a c t}

\Rightarrow F_{c o n t a c t} = 12 N

Commented by Tinkutara last updated on 29/Oct/17

Thank you very much Sir!