Question

There is a uniform meter rod of 100kg, from which a 5kg mass hangs at the 15cm mark, and a 20kg mass hangs from the 70cm mark. At how much distance form the heavier mass should the rod be pivoted?

Answer 1

The rod should be pivoted at a distance of 18.2 cm from the heavier mass.

Answer 2

Let the meter rod be placed along the x-axis, with the 0cm mark at $x=0$ and the 100cm mark at $x=100$ cm.
The rod is uniform with a mass of 100kg. Its weight acts at its center of mass, which is at the midpoint of the rod, i.e., at the 50cm mark. The weight of the rod is $W_{rod} = 100g$, acting downwards at $x_{rod} = 50$ cm.
A mass of $m_1 = 5$ kg hangs at the 15cm mark. Its weight is $W_1 = 5g$, acting downwards at $x_1 = 15$ cm.
A mass of $m_2 = 20$ kg hangs at the 70cm mark. Its weight is $W_2 = 20g$, acting downwards at $x_2 = 70$ cm. This is the heavier mass.

Let the pivot point be located at a distance $x$ from the 0cm mark. For the rod to be in equilibrium, the sum of the torques about the pivot point must be zero.

We can calculate the torque due to each force about the pivot at position $x$. The torque due to a force $F$ acting at position $x_i$ about a pivot at position $x$ is $\tau_i = (x_i - x) F_i$.

The forces are $W_1 = 5g$ at $x_1 = 15$, $W_{rod} = 100g$ at $x_{rod} = 50$, and $W_2 = 20g$ at $x_2 = 70$.
The distances from the pivot at $x$ are $(15-x)$, $(50-x)$, and $(70-x)$.

The torque due to $W_1$ is $(15-x) W_1 = (15-x) 5g$.
The torque due to $W_{rod}$ is $(50-x) W_{rod} = (50-x) 100g$.
The torque due to $W_2$ is $(70-x) W_2 = (70-x) 20g$.

For equilibrium, the sum of the torques is zero: $\sum \tau = (15-x) 5g + (50-x) 100g + (70-x) 20g = 0$

We can divide the equation by $g$: $5(15-x) + 100(50-x) + 20(70-x) = 0$ Expand the terms: $75 - 5x + 5000 - 100x + 1400 - 20x = 0$ Combine the constant terms and the terms with $x$: $(75 + 5000 + 1400) + (-5x - 100x - 20x) = 0$ $6475 - 125x = 0$ $125x = 6475$ $x = \frac{6475}{125} = 51.8$

So, the pivot point is at $x = 51.8$ cm from the 0cm mark.

The question asks for the distance from the heavier mass where the rod should be pivoted. The heavier mass (20kg) is located at the 70cm mark.
The distance from the heavier mass is the absolute difference between the position of the heavier mass and the position of the pivot.
Distance from heavier mass = $|x_{heavier\_mass} - x_{pivot}| = |70 \text{ cm} - 51.8 \text{ cm}| = 18.2 \text{ cm}$.

Thus, the rod should be pivoted at a distance of 18.2 cm from the heavier mass (20kg mass at 70cm mark).