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Two blocks each of mass $m$ are placed on a rough horizontal surface and connected by mass-less in-extensible string as shown.The coefficient of friction between each block and ground is $\mu$ What would be the minimum force $F$ to be applied on the right hand side block to move the block system horizontally without letting the string getting slack.

Now I thought of a simple thing that maximum $$\sum F_{right direction}=F$$ $$\sum F_{left direction}=2\mu mg$$

And hence to just move the block system I need $$F=2\mu mg$$

But my answer is wrong which in turn in $$F=\frac{2\mu mg}{\sqrt{1+\mu^2}}$$

Can anybody give me a hint, where I went wrong.

Qmechanic
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    What would you do if some one asked you to drag a very light object that had a very high friction coefficient? How would you try to move it? – Brian Moths Jun 29 '16 at 22:04
  • I don't see how the force is smaller than your answer, $2\mu m g$. Maybe the "close" voter can give us an answer. – Bill N Jun 29 '16 at 23:12
  • You are obviously paraphrasing a question from a textbook or elsewhere. If the answer is correct then you have missed something important in the question. Please post the original question exactly as written. – sammy gerbil Jun 29 '16 at 23:25
  • Nice question, but did you add something to the drawing that wasn't in the original one? The root of the sum of two squares should also give you some idea of the "direction" you should look to for the solution... – Previous Jun 30 '16 at 00:40
  • May you post the original question with original figure? – lucas Jun 30 '16 at 03:23
  • @lucas I have done it. Please have a look now. –  Jun 30 '16 at 03:54
  • @lucas The figure is the same only as mentioned in the question. –  Jun 30 '16 at 04:00
  • Did my hint do nothing to lift your understanding? – Brian Moths Jun 30 '16 at 05:18
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    @NowIGetToLearnWhatAHeadIs Sorry sir, but I just didn't understand what you were trying to say. –  Jun 30 '16 at 07:48
  • I think the force $F$ shouldn't be exerted horizontally. – lucas Jun 30 '16 at 09:09
  • @lucas Sir can you please explain why? –  Jun 30 '16 at 09:12
  • I ask again, are you sure that the figure is the same original question's figure? – lucas Jun 30 '16 at 09:13
  • It doesn't matter in which direction you think the force should point, the calculation will give you the correct one. Take a horizontal component X, a vertical component Y, calculate the required X in function of Y. Find the value for which F²=X²+Y² is minimal. – Previous Jul 01 '16 at 05:20

2 Answers2

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The original problem wants to find minimum force to move horizontally and this doesn't mean that force $F$ should necessarily be horizontal. enter image description here We have: $$N_1=Mg-F\sin\theta\;\tag 1$$ $$F\cos\theta-T-f_1=0\;\tag 2$$ $$T-f_2=0\;\tag 3$$ $$N_2=Mg\;\tag 4$$ Because of the condition of verge of motion, we have: $$f_1=\mu N_1\;\tag 5$$ $$f_2=\mu N_2\;\tag 6$$ Then, we obtain: $$F=\large{\frac{2\mu Mg}{\cos\theta+\mu\sin\theta}}$$ For minimizing $F$, it should be: $$\large{\frac{\partial F}{\partial\theta}}=0$$ So, it is obtained that: $$\mu=\tan\theta$$ Finally, we reach to the desired answer: $$F_{\textrm{min}}=\frac{2\mu Mg}{\sqrt{1+\mu^2}}$$

lucas
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As you only asked for a hint, not an answer:

Your answer is coming just by considering the force F to be horizontal. But what if it's not horizontal or what if it have a vertical component too? This vertical component will decrease the normal on the right block, so it will also decrease friction.

For minimum force, assume the force to be at some angle, make the equations, differentiate it to get the minimum force, and you will get a particular angle. And for that angle get the force.

Peter Mortensen
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Shubhansh Awasthi
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