# How to find the maximum distance in centimeters so that a sphere supported from one end of a box is at equilibrium?

#### Chemist116

The problem is as follows:

A sphere is placed over a block as seen in the figure from below. The mass of the sphere is $10\,kg$ and the mass of the block is $4\,kg$. Assume that the block is supported over two triangular stands and is aligned horizontaly with respect to the floor. Under these conditions. Find the maximum distance in centimeters from $P$ the sphere can be placed so that the system remains in equilibrium.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{0.4 cm}\\ 2.&\textrm{0.8 cm}\\ 3.&\textrm{1.2 cm}\\ 4.&\textrm{40.0 cm}\\ 5.&\textrm{80.0 cm}\\ \end{array}$

How exactly should I use the equilibrium condition on this problem?. I don't know exactly how to cancel the reaction coming from both supports.

So far the only thing which I could establish (for which I'm not sure if it is correct)

Assuming $N$ is the reaction occuring in both triangular supports

$2N=40+100$

$N=70$

Assuming that the torque happens in the second support seen from the left:

$-N\cdot 3 - 100\cdot (x)+ 40 \cdot 1 = 0$

$-70 \times 3 - 140 x +40=0$

But from solving this system yields a negative quantity. What could I be missunderstanding?. Can someone help me here?. What would be the most appropiate approach here?.

#### skeeter

Math Team
Reference the diagram

$\color{red}C_1,C_2,C_3$ are the center of mass of each section of the block
$\color{blue}x$ is the distance the sphere is from P

Using $\color{green}F_1$ as the pivot for rotational equilibrium ...

$\dfrac{2g}{3} \cdot 0.5 + F_2(3) = 2g \cdot 1.5 + \dfrac{4g}{3} \cdot 4 + 10g \cdot (3+x)$

$F_2 = \dfrac{38g+10gx}{3}$

$F_1 + F_2 = 14g$

$F_1 + \dfrac{g(38+10x)}{3} = 14g \implies F_1 = 14g - \dfrac{g(38+10x)}{3}$

for rotational equilibrium to exist, $F_1 > 0$

$14g - \dfrac{g(38+10x)}{3} > 0$

$42 > 38+10x \implies x<0.4 \text{ m}$

Chemist116

#### Chemist116

Reference the diagram

$\color{red}C_1,C_2,C_3$ are the center of mass of each section of the block
$\color{blue}x$ is the distance the sphere is from P

Using $\color{green}F_1$ as the pivot for rotational equilibrium ...

$\dfrac{2g}{3} \cdot 0.5 + F_2(3) = 2g \cdot 1.5 + \dfrac{4g}{3} \cdot 4 + 10g \cdot (3+x)$

$F_2 = \dfrac{38g+10gx}{3}$

$F_1 + F_2 = 14g$

$F_1 + \dfrac{g(38+10x)}{3} = 14g \implies F_1 = 14g - \dfrac{g(38+10x)}{3}$

for rotational equilibrium to exist, $F_1 > 0$

$14g - \dfrac{g(38+10x)}{3} > 0$

$42 > 38+10x \implies x<0.4 \text{ m}$

View attachment 11088
First things first. Why there are three centers of mass in the block?. Shouldn't it be just one?. Or is it an identity or theorem which am I unaware of?.

For examply why $\frac{2g}{3}$ is the weight at $C_{1}$? How did you obtainted this?.

The other part where I'm stuck at is why for rotational equilibrium to exist means $F_{1}>0$? Shouldn't it be equal to zero?.

Solving the inequality yields $x<0.4\,m$ hence there wouldn't be any answer (according to the listed possibilities). Or could it be that the condition could be $x\leq 0.4\,m$?

Can you please attend these doubts?.

#### skeeter

Math Team
I solved this problem in the manner I did because that's the method I tried first since I saw three unknowns. FYI, one may break up a uniform mass into pieces and determine the center of mass of each piece, where the weight is proportional to the size of each piece.

One could also solve it using the center of mass of 4g N acting downward at the block's center ... I tried this later, and it also worked.

If $x = 0.4 \text{ m} = 40 \text{ cm}$, the force on the left stand becomes zero and the block is at the tipping point. The problem asked for a maximum ... if you want to make the inequality less than or equal to, then go for it.

#### Chemist116

I solved this problem in the manner I did because that's the method I tried first since I saw three unknowns. FYI, one may break up a uniform mass into pieces and determine the center of mass of each piece, where the weight is proportional to the size of each piece.

One could also solve it using the center of mass of 4g N acting downward at the block's center ... I tried this later, and it also worked.

If $x = 0.4 \text{ m} = 40 \text{ cm}$, the force on the left stand becomes zero and the block is at the tipping point. The problem asked for a maximum ... if you want to make the inequality less than or equal to, then go for it.
As you indicated I overlooked the fact that you can split a big object into small pieces and assign to each one a weight proportial to that size. (Would it work in non uniform objects too?).
So If I follow this logic then:

$C_{1}=\frac{1}{6}4g=\frac{2}{3}g$ and so on for the others.

But If I'm understanding this correctly. If you use the center of mass 4g acting at the block's center, does it means in that scenario $C_{1}$ and $C_{3}$ will not exist?. Or $C_{1}$ and $C_{3}$ must appear in the equation for equilibrium?.

If I were to use only the weight acting in the center of mass I'm getting:

$F_{2}\cdot 3=4g\cdot 2 +10g\cdot (3+x)$

$F_{2}=\frac{10gx+38g}{3}$

Going even further it will produce:

$42\geq 38+10x$ and yield the same result confirming what you mentioned.

There is an unatended doubt here which isn't yet answered. Why $F_{1}>0$ for rotational equilibrium here?. I thought that the rotational equilibrium to exist means that the net torque is equal to zero. Which you already solved in the first equation for $F_{1}$. Can you better explain this part?. I'm assuming that the reason is that when the force at the support is zero (from what you mentioned) then the object is about to tip over, but not yet. Then it would make sense that $x$ will be at its maximum. Am I right with this interpretation?.

I'm also trying to understand what if the force is lesser than zero?. What would be the interpretation for this?. Would it be that the object is "floating" about that point?. Can you attend these questions please?.

#### skeeter

Math Team
$F_1$ can only be greater than or equal to zero. It cannot be less than zero. If at zero, the block is at the point of tipping. If the sphere were moved farther right from the tipping point, $F_1$ is still zero.