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Today when I saw my phone on table some of its part was off the table, so I wondered what will be the arrangement of phone to balance it on corner of table such that it's minimum area (of screen) is on the table?

Assume that density of phone is uniform and it has dimension of $l× b$.

My attempt: In first scenario let corner of table be perpendicular to phone and we move phone little by little and see balancing point . Intuition tells me answer is gonna be ${l\over2}$ or ${b\over2}$ . But in proof I'm getting some confusions

  1. Some part of phone is off the table so normal force is still gonna be $mg$ but I can't get feel that how this normal force is gonna distribute over part that is on the table ? Is it uniform ?

  2. What if the part of phone that is off table has no uniform mass distribution then how normal force distribute over part on the table or on very small element ( dx.dy ) of phone ?

  3. How we know that there is or not any rotational orientation case of phone which gave us area less then $lb\over2$ ?

  4. What if we include part of table where two corner sides meet ? In this case I think area of one or two right angles will be on the table .

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  • $\begingroup$ I would have tried to keep the phone vertical such that the point of contact is just below the center of gravity(although it will be in unstable equilibrium but you just asked for equilibrium). $\endgroup$
    – Ankit
    Nov 9, 2021 at 11:18

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Look at the phone (red)/table (black) system from above:

Eagle's eye

(I've assumed that the angle between $|PE|$ and the side of the table is $45^{\circ}$)

Note the significant points $P$, $E$ and $C$.

A sideways glance:

Sideways glance

In order for the phone to be motionless, per Newton:

  • net force acting on phone must be zero
  • net torque acting on phone must be zero

The first condition is fulfilled because of $\text{N3L}$:

$$R=mg$$

For the second condition we take the torque balance about the point $E$.

If:

$$m\vec{g} \times |CE|>0\text{ (clockwise torque)}$$

then the phone will topple, because there is net, clockwise torque causing angular acceleration.

If:

$$m\vec{g} \times |CE|<0\text{ (anti-clockwise torque)}$$

then it will remain stable.

In other words, $C$ must fall between $P$ and $E$ for stability to occur.

So in the case of a uniform phone, your answer of $\frac{l}{2}$ or $\frac{b}{2}$ (depending on orientation) is correct.

Regards $1.$, at the tipping point, $R$ is concentrated in $E$.

Regards $2$, non-uniformity will affect the position of $C$ but not the above reasoning.

Point $3.$ has been answered.

But sorry: I really don't get your point $4.$ Diagram maybe?

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  • $\begingroup$ Does $R$ act all throughout the phone or only at $C$? $\endgroup$ Nov 9, 2021 at 9:53
  • $\begingroup$ Hi. It depends. When the phone is just about to tip down, $R$ only acts on $E$ but when $C$ is well between $P$ and $E$, $R$ acts uniformly between $P$ and $E$. Does that help? $\endgroup$
    – Gert
    Nov 9, 2021 at 11:08
  • $\begingroup$ By "just about to tip down", do you mean when the $C$ is slightly to the right of $E$? Also, where does it act when the $C$ is on $E$? Is it $C$ and $E$ at the same time? And also what about when $C$ is considerably to the right of $E$, as shown by you in the diagram? $\endgroup$ Nov 9, 2021 at 11:53
  • $\begingroup$ @Gert Okay I get it when phone is about to tip down normal is concentrated at point (line )C . But what if we move phone little more on table how will normal force distribution affect ? Is it on some axis or point or uniform ? Does it depends on distribution of mass density on or off the table ? $\endgroup$
    – RKK
    Nov 9, 2021 at 15:37
  • $\begingroup$ @RKK When $C$ is well between $P$ and $E$, the back of the phone is firmly in contact with the table surface, so $R$ is uniformly distributed over that surface area. $\endgroup$
    – Gert
    Nov 9, 2021 at 18:19

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