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enter image description here

A $5$ kg and a $10$ kg box are touching each other. A $45N$ horizontal force is applied to the $5$kg box in order to accelerate both boxes across the floor. Ignore friction forces and determine the force between the hand and the boxes.

Considering the body as a single mass, we can find the acceleration of the body as $3\ \mathrm{ms}^{-2}$. The force between the blocks can similarly be found as $30\ \mathrm{N}$. But coming to the question, I cannot be sure as to whether to take the two blocks as a single body. If I do so, then as per my understanding of Newton's Third Law, the hand should experience a force of $-45\ \mathrm{N}$. Now, despite my hand experiencing an equal force in the opposite direction, I am able to keep contact with the block and continue pushing the 2 blocks over a distance (say $d$). Could someone please help me by explaining why this is so? Thanks a lot!

(This is not an original question. I had come across it 'the Physics Classroom' and thought of this variant.)

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  • $\begingroup$ What would be the problem with having a negative force? You say you exert a force of $45N$ on the box so surely you would expect by $N3L$ that the box would exert the same force in the opposite direction onto your hand? $\endgroup$ – tey yreryt May 1 '15 at 15:43
  • $\begingroup$ Sorry, I somehow forgot to add a few lines to the question. I'm editing it now. Please do help. $\endgroup$ – WORLD 1 May 1 '15 at 15:46
  • $\begingroup$ @teyyreyt Is it alright now? $\endgroup$ – WORLD 1 May 1 '15 at 15:50
  • $\begingroup$ I believe (could be wrong) that you are asking why would the boxes keep moving when there is a force of $45N$ acting in opposite directions so why don't they balance and the boxes don't move at all. Well when you are looking at movement and forces you need to consider that forces ACTING ON THE BOXES i.e. the 45N rather than the forces acting on the hand. So just observe the boxes there is a 45N force hence the boxes move. The reaction force is irrelevant to the motion as that is acting on the hand not onto the boxes. $\endgroup$ – tey yreryt May 1 '15 at 15:53
  • $\begingroup$ Granted, but then why should my hand also keep moving in contact with the boxes? Shouldn't the hand in fact recoil backwards? $\endgroup$ – WORLD 1 May 1 '15 at 15:55
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First We'll see the FBD (Free Body Diagram)and we get: enter image description here

Where Fc is normal force acting (Force of contact).

From FBD of 5 Kg block (By newton's 2nd law)

$$F-F_c = ma$$ (1)

From FBD of 10 Kg block

$$F_c=Ma$$

Solving above equations we will get:

$$F=ma+Ma$$ $$a=F/(m+M)$$

Putting the values you may get your result and your resultant force.

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  • $\begingroup$ $F-F_c = ma$ .$F_c=Ma$. From these, shouldn't we get $F=ma+Ma$? $\endgroup$ – WORLD 1 May 1 '15 at 16:05
  • $\begingroup$ @WORLD1 You are right ... Just get it little bit wrong :P i have edited $\endgroup$ – Shashank May 1 '15 at 16:12
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The net force on the 5kg block will be 15N in the direction of the hand pushing. the subsequent acceleration or the 5kg block will be 3m per second per second. the net force on the 10kg block will be 30N giving an acceleration 3m per second per second. The accelleration will continue until the the hand is removed resulting in the blocks continuing at a constant velocity until a further force is actioned on the boxes.

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  • $\begingroup$ But why should the hand be able to keep on moving in contact with the blocks? Shouldn't Newton's Third Law prevent it from doing so? $\endgroup$ – WORLD 1 May 1 '15 at 16:29
  • $\begingroup$ I Think yes the hand does experience a force of 45N in the opposite direction meaning the relative velocity of the hand and the boxes remains zero. But the hand has to accellerate along with the boxes in order to maintain the zero relative velocity between them. $\endgroup$ – 8Mad0Manc8 May 1 '15 at 16:46
  • $\begingroup$ Could you please explain how the relative velocity between the hand and the boxes remains $0$? Thanks a lot in advance! $\endgroup$ – WORLD 1 May 1 '15 at 17:03
  • $\begingroup$ If the hand and the box are accelerating at the same amount and in the same direction the relative velocity between the two is zero ie they are travelling at the same velocity in the same direction at any instant in time thus the relative velocity between them is zero. Iam still thinking about your question though. $\endgroup$ – 8Mad0Manc8 May 1 '15 at 17:08
  • $\begingroup$ I must admit newtons third law states equal and opposite thus if a 45N force is exerted in one direction the same must be exerted in the other but if this was the case every force would result in no change in an objects acceleration. $\endgroup$ – 8Mad0Manc8 May 1 '15 at 17:35

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