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I was doing the problem described in the following question:

A $0.250\ \mathrm{kg}$ block of cheese lies on the floor of a $900\ \mathrm{kg}$ elevator cab that is being pulled upward by a cable through distance $d_1=2.40\ \mathrm{m}$ and then through distance $d_2= 10.5\ \mathrm{m}$. (a) Through $d_1$, if the normal force on the block from the floor has constant magnitude $F_N= 3.00\ \mathrm{N}$, how much work is done on the cab by the force from the cable? (b) Through $d_2$, if the work done on the cab by the (constant) force from the cable is $92.61\ \mathrm{kJ}$, what is the magnitude of $F_N$?

And had the same issue with seeing around the Internet a lot of solutions involving the normal force of the box when it comes to calculating the tension in this type of exercise

Before finding the question linked above that suggests that the normal force shouldn't be involved, I tried to convince myself that using it was wrong.

My reasoning is below and I'd appreciate knowing if it was correct.

Taking the same situation of the exercise, in general terms, I'm going to assume that the solutions involving a normal force of the box are correct:

In the first place, the normal force acting on the box is given by:

$$ N - m_{b}g = m_{b}a $$ $$ N = m_{b}a + m_{b}g $$ $$ N = m_{b}(a + g) $$

To calculate the Tension force, using the 2nd Newton's law and looking at the forces acting on the elevator + box system:

$$ T + N - (m_{b}+m_{e})g = (m_{b}+m_{e})a $$ $$ T = (m_{b}+m_{e})a + (m_{b}+m_{e})g - N $$ $$ T = (m_{b}+m_{e})(a + g) - N $$ $$ T = (m_{b}+m_{e})(a + g) - m_{b}(a + g) $$ $$ T = (m_{b}+m_{e}- m_{b})(a + g) $$ $$ T = m_{e}(a + g) $$

So apparently, this is telling me that the tension in the cable of the elevator doesn't get affected by whatever bodies with mass are inside the elevator. Which intuitively I think is wrong.

Would this reasoning be valid?

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$$ T + N - (m_{b}+m_{e})g = (m_{b}+m_{e})a $$ is wrong, for two reasons. First, you said you are applying Newton's second law to the elevator (I'm assuming just the cab, without the cheese). Then, the mass in Newton's second law is just that of the elevator cab.

The second problem is the sign of $N$. It pushes the elevator cab downward (the direction opposite to that of the acceleration), so it must appear on the left hand side with a negative sign. The correct equation is $$ T - N - m_{e}g = m_{e}a. $$

If you are simply looking to relate $T$ and $a$, you can apply Newton's second law to the combined cab and cheese system. The normal force is an internal force in this system and has no effect on the acceleration. Newton's second law reads $$ T - (m_b + m_e)g = (m_b + m_e)a. $$

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  • $\begingroup$ Is the normal force you are mentioning the weight of the box in the elevator? In my example, I'm using the normal force of the box, as it is an example to say why using that internal normal force is wrong $\endgroup$
    – Jon
    Commented Oct 30, 2020 at 19:00
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    $\begingroup$ The normal force is different from the weight of the box. The normal force on the box is the contact force applied by the elevator on the box, and the normal force on the elevator is the contact force applied by the box on the elevator. By Newton's third law, these forces are equal in magnitude and opposite in direction, and you don't consider them only if the system you are applying Newton's second law to is the combined elevator + block system since the sum of these two forces is zero. $\endgroup$
    – Puk
    Commented Oct 30, 2020 at 19:05
  • $\begingroup$ Thanks. From that perspective is much more clear. $\endgroup$
    – Jon
    Commented Oct 30, 2020 at 19:12

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