# What is the Acceleration?

Find the acceleration of the $$2\text{ kg}$$ block in $$\text{ms}^{-2}$$ ?

###### Source: This beautiful question appeared at eagerbug.com . I hope everyone like solving it.! :)

Note by Rishabh Tiwari
5 years, 1 month ago

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Let me revise frictional force topic again has been two years since I last learned it ;)

- 5 years, 1 month ago

As you say my friend:-)

- 5 years, 1 month ago

If those are static friction coefficients, then $a = 0$ is possible.
If these are kinetic friction coefficients, write the force balance for each block. I'm assuming the wedge is massless.
The centre of mass will move downwards only, so acceleration of the heavier block will be down the plane, and the lighter one will move up the plane. As the string does not extend/contract, these accelerations are same in magnitude.
Can you solve it from here?

- 5 years, 1 month ago

Right sir, thank you.!

- 5 years, 1 month ago

- 5 years, 1 month ago

Does that mean you won't post a solution? Please do! I need to understand it completely!

- 5 years, 1 month ago

Taking the 3kg block accelerating downwards $T-\mu mg\cos { \theta } -mg\sin { \theta } =ma\quad \quad \quad \quad \quad \quad \quad \quad \quad \left( 1 \right) \\ Mg\sin { \alpha } -\mu 'Mg\cos { \alpha } =Ma\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left( 2 \right) \\ Adding\quad 1\quad and\quad 2\quad we\quad get\\ \left( M\sin { \alpha } -m\sin { \theta } \right) g-g\left( \mu 'M\cos { \alpha } +\mu m\cos { \theta } \right) =\left( M+m \right) a\\ \Rightarrow 18-16-4.8-4.8=5a\\ \Rightarrow a=-\frac { 7.2.2 }{ 5 } =-1.52m{ s }^{ -2 }\\ \\$ - sign means that the acceleration of the system is in opposite direction to that I have assumed here, i.e. 2kg block slides downward. However taking the other block(i.e. 2kg) accelerating downwards, $-18+16-4.8-4.8=5a\\ \Rightarrow a=-\frac { 11.6 }{ 5 } =-2.32m{ s }^{ -2 }\\$ As these results are contradictory, I think that the system wont accelerate. So a=0

- 5 years, 1 month ago

- 5 years, 1 month ago

@Swapnil Das ; a good problem for you & for the physics section, I hope ! :-)

- 5 years, 1 month ago

@Ashish Siva, @Deeparaj Bhat , @Abhay Tiwari , please comment!

- 5 years, 1 month ago

Sorry Rishabh, I am not good at this topic.

- 5 years, 1 month ago

Not a problem bro, Ur Perfect! Me2 not good at this.

- 5 years, 1 month ago

I am not sure but I guess the 3kg will pull with a Force of $F=g[(3 \space Sin (37°)-0.2×3 ×\space Cos(37°))\space - \space (2 \space Sin (53°)-0.4×2× \space Cos(53°))]=2.1481 \space kg \space ms^{-2}$, if $g=10 \space ms°^{-2}$

- 5 years, 1 month ago

I think the second half of the force is not needed, first half would be the force driving the $3kg$ block down & you also have to subtract tension from that!

- 5 years, 1 month ago

Ha ha. Let me see if somebody posts a solution. Maybe then I will understand.

- 5 years, 1 month ago

Ya right, I expect a good solution from @Swapnil Das or @Deeparaj Bhat , however, anyone is free to post a solution! :-)

- 5 years, 1 month ago

Hey does anyone know the answer? I got $\boxed{1.52 m/s^2}$

- 5 years, 1 month ago

Also according to my solution the 2 kg block moves up the slope @Rishabh Tiwari Do you know the correct answer ?

- 5 years, 1 month ago

I don't know the answer but here are the options:

(a) 2 (b) 0

(c) 2.32 (d) 1.8

- 5 years, 1 month ago