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Case 1:

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Three blocks are being pushed by the force $F$. The horizontal surface and surface between blocks 1 and 2 are smooth. Only the surface between blocks 2 and 3 is rough. We did not consider the normal force between $m_1$ and $m_2$, as my teacher said that we should only consider the normal between the required surfaces.

Case 2:

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Here, we have two blocks on top of each other, so according to his logic, to find friction between ground and block, we should not consider the force of upper block on the lower block as it is not in direct contact, but I think that should not be the case.

Please explain the concept as I am very confused.

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Conceptually whenever we apply the friction force ($f=\mu N$), the $N$ is the force acting on the surface and not on the block, for example, a block of mass $m$ kept on a rough surface, then for equilibrium of body $$mg=N$$ and as a result of Newton's Third Law, $N$ acts on ground, therefore the net force on the friction surface/ground is $N$ and the normal in friction is basically this $N$. This fundamental concept is very important in solving the questions. So in your case 1,
let's say we draw individual FBD's, so on $m_2$ the force from right block is $N_2$, which is shown in the diagram. Therefore, the force on surface of $m_2, m_3$ is $N_2$, hence $f_s=\mu N_2.$

And for case 2,
draw the individual FBD's step-by-step, as done above so the normal force on ground is $$N=(m_1+m_2)g$$
Therefore, whenever we want to apply the friction force, always consider the net force acting on the friction surface, accordingly to the opposite of that force will be Normal Force which we substitute in the equation of friction. Hope this clears your confusion.

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  • $\begingroup$ so only m3 is providing normal to m2 , so what about the force that m1 applies on m2, is it just pushing it forward for accleration and not contributing to normal? Please help $\endgroup$ Jul 17 '21 at 10:01
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    $\begingroup$ Yes it is contributing to acceleration of $m_2$ but as I told in the answer, we have to see what force acts on the rough surface, i.e, only $N_2$, I'll advise you to draw a diagram manually on paper, ofcourse $N_1$ is contributing normal to $m_2$, but what we see is what normal does the surface apply on the block in contact, and then the opposite is the reaction force $N_2$ on surface therefore, $\mu N_2$ $\endgroup$
    – PCMSE
    Jul 17 '21 at 10:05
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    $\begingroup$ @AkshatJangra, Now, clear? Actually the diagram itself gives what you are questioning about $\endgroup$
    – PCMSE
    Jul 17 '21 at 10:10
  • $\begingroup$ absolutely clear now thanks $\endgroup$ Jul 17 '21 at 10:56
  • $\begingroup$ @AkshatJangra, Great $\endgroup$
    – PCMSE
    Jul 17 '21 at 11:13
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First we see why the normal force arises, when we push against a surface, the surface pushes back against us with a equal and opposite force by Newton's third law. This force is the normal force. Friction is proportional to this normal force.

In both cases the normal forces of the required surface have to be considered. The only difference in case 2 is that the body has zero acceleration downwards, so the normal force must be equal to the weight of both the bodies. If that was not the case, the body would go through the ground

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