# 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
4 years, 11 months ago

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

- 4 years, 11 months ago

As you say my friend:-)

- 4 years, 11 months 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?

- 4 years, 11 months ago

Right sir, thank you.!

- 4 years, 11 months ago

- 4 years, 11 months ago

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

- 4 years, 11 months 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

- 4 years, 10 months ago

- 4 years, 10 months ago

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

- 4 years, 11 months ago

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

- 4 years, 11 months ago

Sorry Rishabh, I am not good at this topic.

- 4 years, 11 months ago

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

- 4 years, 11 months 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 $g=10 \space ms°^{-2}$

- 4 years, 11 months 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!

- 4 years, 11 months ago

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

- 4 years, 11 months ago

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

- 4 years, 11 months ago

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

- 4 years, 10 months ago

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

- 4 years, 10 months ago

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

(a) 2 (b) 0

(c) 2.32 (d) 1.8

- 4 years, 10 months ago