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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
1 year, 5 months ago

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

Ashish Siva - 1 year, 5 months ago

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As you say my friend:-)

Rishabh Tiwari - 1 year, 5 months ago

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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?

Ameya Daigavane - 1 year, 5 months ago

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Right sir, thank you.!

Rishabh Tiwari - 1 year, 5 months ago

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I guess Ameya already answered it. ;)

Deeparaj Bhat - 1 year, 5 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 - 1 year, 5 months ago

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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

Swagat Panda - 1 year, 5 months ago

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

Rishabh Tiwari - 1 year, 5 months ago

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

Abc Xyz - 1 year, 5 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 - 1 year, 5 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 - 1 year, 5 months ago

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Comment deleted Jun 27, 2016

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Friction?

Ameya Daigavane - 1 year, 5 months ago

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Both frictional forces have been included, However, some errors have crept into the calculation. I'd request someone to provide us with the correct calculation. As a matter of fact, I now understand what's happening here. I calculated the acceleration both ways and it so happens that it turns out to be negative each time. This means that frictional forces cannot be overcome and, therefore, the blocks will remain stationary or a=0.

One Top - 1 year, 5 months ago

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@One Top I think your answer is correct. An option is there of 0.

Rishabh Tiwari - 1 year, 5 months ago

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As Ameya asked, you also need to use the frictional force while solving.

Rishabh Tiwari - 1 year, 5 months ago

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

Rishabh Tiwari - 1 year, 5 months ago

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Sorry Rishabh, I am not good at this topic.

Abhay Tiwari - 1 year, 5 months ago

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

Rishabh Tiwari - 1 year, 5 months ago

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

Abhay Tiwari - 1 year, 5 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 \(3kg\) block down & you also have to subtract tension from that!

Rishabh Tiwari - 1 year, 5 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 - 1 year, 5 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 - 1 year, 5 months ago

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

Rishabh Tiwari - 1 year, 5 months ago

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