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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.)

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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.