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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.)

Community
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WORLD 1
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  • 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? – tey yreryt May 01 '15 at 15:43
  • Sorry, I somehow forgot to add a few lines to the question. I'm editing it now. Please do help. – WORLD 1 May 01 '15 at 15:46
  • @teyyreyt Is it alright now? – WORLD 1 May 01 '15 at 15:50
  • 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. – tey yreryt May 01 '15 at 15:53
  • Granted, but then why should my hand also keep moving in contact with the boxes? Shouldn't the hand in fact recoil backwards? – WORLD 1 May 01 '15 at 15:55
  • I would add to @teyyreryt the following: your hand isn't a box. If it was, the box wouldn't feel any force. When you push a box, all of time you push it you feel force. That's because your hand pushes with a greater force than $45N$. (correct me if I'm wrong) – Nadav S. May 01 '15 at 15:57
  • @WORLD1 You've analyzed all the forces on the boxes, but you haven't analyzed all the forces on the hand. I think you'll find that an arm is providing a force on it (probably around 45N or so) – BowlOfRed May 01 '15 at 15:57
  • True, but then could this not extended to the arm? If the arm is exerting a force, should it not feel an equal reaction force in the opposite direction which would neutralize the force the arm exerts? – WORLD 1 May 01 '15 at 16:02
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    Possible duplicates: https://physics.stackexchange.com/q/45653/2451 and links therein. – Qmechanic Apr 14 '17 at 20:43
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    Does this answer your question? Given Newton's third law, why are things capable of moving? – Brian Drake Nov 15 '20 at 13:11

2 Answers2

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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.

Shashank
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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.

8Mad0Manc8
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  • 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? – WORLD 1 May 01 '15 at 16:29
  • 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. – 8Mad0Manc8 May 01 '15 at 16:46
  • Could you please explain how the relative velocity between the hand and the boxes remains $0$? Thanks a lot in advance! – WORLD 1 May 01 '15 at 17:03
  • 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. – 8Mad0Manc8 May 01 '15 at 17:08
  • 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. – 8Mad0Manc8 May 01 '15 at 17:35
  • Not necessarily, Sir. Assume that the hand was to push the box with $10N$. The box would then move forward (assuming the force was enough to overcome friction). However, the hand should then stop exerting any force and become stationary instead of moving with the box... – WORLD 1 May 01 '15 at 17:46
  • Yes i think your getting into further issues like impulses, inertia and friction. Friction is dependent upon the vertical force of gravity and the constant component of friction between the two that is dependent on the objects that are in contact. For the case you have described you should ignore friction and inertia – 8Mad0Manc8 May 01 '15 at 18:04