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

Question

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?.

Answer 1

Let $P$ be the pivot and let $d$ be the maximum distance of the sphere. If the sphere is maximally far to the right, then the reaction force in the left support should be zero. So we just need to balance the torque of the block with the torque of the sphere. Assuming that the block is uniform, we have: $$ (4 \text{ kg})(g \text{ m/s$^2$})(1 \text{ m}) = (10 \text{ kg})(g \text{ m/s$^2$})(d \text{ m}) \iff d = 0.4 \text{ m} = 40 \text{ cm} $$

Answer 2

When I solve physics problems I usually try to consider the general case and come up with general formulas for a certain type of problem. Besides being extremely fun, this is very instructive since it teaches you to analyze general situations abstractly using symbols and improves your ability to identify separate cases of a certain situation, so here it is:

Suppose we have a block of uniformly distributed mass $M$ and length $L$ and suppose it is supported by two triangular stands set at positions $b$ and $a$ from left to right respectively where the origin is taken to be the leftmost end of the block. Define the positive rotation direction to be counterclockwise. Let $m$ be the mass of the sphere and suppose it is put at a distance $p$ from our origin such that $p>a>L/2>b$. Now the center of mass of the block is $L/2$ before we put the sphere and we consider the actions of the forces with respect to this center of mass. Let $n_a$ and $n_b$ be the normal forces at $a$ and $b$ respectively. Using the conditions for equilibrium we get: $$n_{a}+n_{b}=g\left(M+m\right)$$

$$\left(a-\frac{L}{2}\right)n_{a}-\left(\frac{L}{2}-b\right)n_{b}=gm\left(p-\frac{L}{2}\right)$$ Solving them simultaneously for $n_a$ and $n_b$ gives: $$n_{a}=\frac{g\left(M+m\right)\left(\frac{L}{2}-b\right)}{a-b}+\frac{gm\left(p-\frac{L}{2}\right)}{a-b}$$ $$n_{b}=\frac{g\left(M+m\right)\left(a-\frac{L}{2}\right)}{a-b}-\frac{gm\left(p-\frac{L}{2}\right)}{a-b}$$ Now note that both forces, by newton's third law, must point upwards which means that they must be positive in our coordinate system. You may note that $n_a$ is positive for all values of $p$ and it is an increasing function of $p$. However, $n_b$ is a decreasing function of $p$ and since it can't be negative, it is the "limiting factor" in some sense. Suppose we could find a $p_{m}$ such that $n_b =0$, this gives: $$p_{m}=\frac{\left(M+m\right)\left(a-\frac{L}{2}\right)}{m}+\frac{L}{2}$$ Note that $p_{m}>\left(a-\frac{L}{2}\right)-\frac{L}{2}=a$ since the minimal value of $\frac{M+m}{m}$ is $1$. Suppose the maximum distance $p_{max}$ was attained somewhere between $a$ and $p_{m}$, this is a contradiction since the above equations show that there is some $p$ such that $p_{max}<p< p_m$ such that $n_b$ and $n_a$ are positive and the system is in equilibrium. This shows that $p_{max}\ge p_m$, but it can't be larger than $p_{m}$ since that would make $n_b$ negative. We conclude that $p_m=p_{max}$. Now the answer to your question requires the distance from $a$ to $p_{max}$ and not $p_{max}$ itself, it also wants it in centimeters. So the general solution to your problem is this:

$$p_{max}=100\left(\frac{\left(M+m\right)\left(a-\frac{L}{2}\right)}{m}+\frac{L}{2}-a\right)$$

You can check that it gives the answer $40cm$ in your specific case which confirms Adriano's answer.