What is the Acceleration?

Find the acceleration of the 2 kg2\text{ kg} block in ms2\text{ms}^{-2} ?


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

Note by Rishabh Tiwari
3 years, 4 months ago

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

Ashish Siva - 3 years, 4 months ago

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As you say my friend:-)

Rishabh Tiwari - 3 years, 4 months ago

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If those are static friction coefficients, then a=0 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?

Ameya Daigavane - 3 years, 4 months ago

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Right sir, thank you.!

Rishabh Tiwari - 3 years, 4 months ago

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I guess Ameya already answered it. ;)

Deeparaj Bhat - 3 years, 4 months ago

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

Rishabh Tiwari - 3 years, 4 months ago

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Taking the 3kg block accelerating downwards Tμmgcosθmgsinθ=ma(1)MgsinαμMgcosα=Ma(2)Adding1and2weget(Msinαmsinθ)gg(μMcosα+μmcosθ)=(M+m)a18164.84.8=5aa=7.2.25=1.52ms2T-\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+164.84.8=5aa=11.65=2.32ms2-18+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

Swagat Panda - 3 years, 4 months ago

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

Rishabh Tiwari - 3 years, 4 months ago

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

Rishabh Tiwari - 3 years, 4 months ago

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

Rishabh Tiwari - 3 years, 4 months ago

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Sorry Rishabh, I am not good at this topic.

Abhay Tiwari - 3 years, 4 months ago

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

Rishabh Tiwari - 3 years, 4 months ago

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@Rishabh Tiwari I am not sure but I guess the 3kg will pull with a Force of F=g[(3 Sin(37°)0.2×3× Cos(37°))  (2 Sin(53°)0.4×2× Cos(53°))]=2.1481 kg ms2F=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 ms°2g=10 \space ms°^{-2}

Abhay Tiwari - 3 years, 4 months ago

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@Abhay Tiwari I think the second half of the force is not needed, first half would be the force driving the 3kg3kg block down & you also have to subtract tension from that!

Rishabh Tiwari - 3 years, 4 months ago

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

Abhay Tiwari - 3 years, 4 months ago

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

Rishabh Tiwari - 3 years, 4 months ago

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Hey does anyone know the answer? I got 1.52m/s2\boxed{1.52 m/s^2}

abc xyz - 3 years, 4 months ago

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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 ?

abc xyz - 3 years, 4 months ago

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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

Rishabh Tiwari - 3 years, 4 months ago

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