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

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

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TopNewestLet me revise frictional force topic again has been two years since I last learned it ;) – Ashish Siva · 11 months, 2 weeks ago

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– Rishabh Tiwari · 11 months, 2 weeks ago

As you say my friend:-)Log in to reply

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 · 11 months, 2 weeks ago

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– Rishabh Tiwari · 11 months, 2 weeks ago

Right sir, thank you.!Log in to reply

– Deeparaj Bhat · 11 months, 2 weeks ago

I guess Ameya already answered it. ;)Log in to reply

– Rishabh Tiwari · 11 months, 2 weeks ago

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

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 · 11 months, 2 weeks ago

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– Rishabh Tiwari · 11 months, 2 weeks ago

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

Hey does anyone know the answer? I got \(\boxed{1.52 m/s^2}\) – Abc Xyz · 11 months, 2 weeks ago

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(a) 2 (b) 0

(c) 2.32 (d) 1.8 – Rishabh Tiwari · 11 months, 2 weeks ago

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@Rishabh Tiwari Do you know the correct answer ? – Abc Xyz · 11 months, 2 weeks ago

Also according to my solution the 2 kg block moves up the slopeLog in to reply

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– Ameya Daigavane · 11 months, 2 weeks ago

Friction?Log in to reply

– One Top · 11 months, 2 weeks ago

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.Log in to reply

– Rishabh Tiwari · 11 months, 2 weeks ago

I think your answer is correct. An option is there of 0.Log in to reply

– Rishabh Tiwari · 11 months, 2 weeks ago

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

@Ashish Siva, @Deeparaj Bhat , @Abhay Tiwari , please comment! – Rishabh Tiwari · 11 months, 2 weeks ago

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– Abhay Tiwari · 11 months, 2 weeks ago

Sorry Rishabh, I am not good at this topic.Log in to reply

– Rishabh Tiwari · 11 months, 2 weeks ago

Not a problem bro, Ur Perfect! Me2 not good at this.Log in to reply

– Abhay Tiwari · 11 months, 2 weeks 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 \(g=10 \space ms°^{-2}\)Log in to reply

– Rishabh Tiwari · 11 months, 2 weeks 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!Log in to reply

– Abhay Tiwari · 11 months, 2 weeks ago

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

@Swapnil Das or @Deeparaj Bhat , however, anyone is free to post a solution! :-) – Rishabh Tiwari · 11 months, 2 weeks ago

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

@Swapnil Das ; a good problem for you & for the physics section, I hope ! :-) – Rishabh Tiwari · 11 months, 2 weeks ago

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