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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
3 months, 2 weeks ago

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Let me revise frictional force topic again has been two years since I last learned it ;) Ashish Siva · 3 months, 2 weeks ago

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@Ashish Siva As you say my friend:-) Rishabh Tiwari · 3 months, 2 weeks 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 · 3 months, 2 weeks ago

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@Ameya Daigavane Right sir, thank you.! Rishabh Tiwari · 3 months, 2 weeks ago

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@Rishabh Tiwari I guess Ameya already answered it. ;) Deeparaj Bhat · 3 months, 2 weeks 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 months, 2 weeks 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 · 3 months, 1 week ago

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@Swagat Panda Please retry and calculate and see if your answer matches with the options I have provided. Rishabh Tiwari · 3 months, 1 week ago

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Hey does anyone know the answer? I got \(\boxed{1.52 m/s^2}\) Abc Xyz · 3 months, 1 week ago

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

(a) 2 (b) 0

(c) 2.32 (d) 1.8 Rishabh Tiwari · 3 months, 1 week ago

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@Abc Xyz Also according to my solution the 2 kg block moves up the slope @Rishabh Tiwari Do you know the correct answer ? Abc Xyz · 3 months, 1 week ago

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@One Top Friction? Ameya Daigavane · 3 months, 1 week ago

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@Ameya Daigavane 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 · 3 months, 1 week ago

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@One Top I think your answer is correct. An option is there of 0. Rishabh Tiwari · 3 months, 1 week ago

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@One Top As Ameya asked, you also need to use the frictional force while solving. Rishabh Tiwari · 3 months, 1 week ago

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@Ashish Siva, @Deeparaj Bhat , @Abhay Tiwari , please comment! Rishabh Tiwari · 3 months, 2 weeks ago

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@Rishabh Tiwari Sorry Rishabh, I am not good at this topic. Abhay Tiwari · 3 months, 2 weeks ago

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@Abhay Tiwari Not a problem bro, Ur Perfect! Me2 not good at this. Rishabh Tiwari · 3 months, 2 weeks 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 · 3 months, 2 weeks 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 · 3 months, 2 weeks 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 months, 2 weeks 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 months, 2 weeks ago

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@Swapnil Das ; a good problem for you & for the physics section, I hope ! :-) Rishabh Tiwari · 3 months, 2 weeks ago

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