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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?

@Rishabh Tiwari
–
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 $g=10 \space ms°^{-2}$

@Abhay Tiwari
–
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!

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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

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As you say my friend:-)

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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?

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Right sir, thank you.!

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I guess Ameya already answered it. ;)

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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

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Please retry and calculate and see if your answer matches with the options I have provided.

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

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

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Sorry Rishabh, I am not good at this topic.

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

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$F=g[(3 \space Sin (37°)-0.2×3 ×\space Cos(37°))\space - \space (2 \space Sin (53°)-0.4×2× \space Cos(53°))]=2.1481 \space kg \space ms^{-2}$, if $g=10 \space ms°^{-2}$

I am not sure but I guess the 3kg will pull with a Force ofLog in to reply

$3kg$ block down & you also have to subtract tension from that!

I think the second half of the force is not needed, first half would be the force driving theLog in to reply

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@Swapnil Das or @Deeparaj Bhat , however, anyone is free to post a solution! :-)

Ya right, I expect a good solution fromLog in to reply

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

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

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

(a) 2 (b) 0

(c) 2.32 (d) 1.8

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