# Question #12ba5

Jun 9, 2016

Since the selection of answers is not given, we will have to find a relation which a possible answer must satisfy.

If the system is stable, the net torque must be equal. The 4 kg mass on the left end exerts a torque of:
${\tau}_{\text{left}} = F d = M \cdot g \cdot d = 4 k g \cdot 10 \frac{m}{s} ^ 2 \cdot 4.5 m$
${\tau}_{\text{left}} = 190 \frac{k g \cdot {m}^{2}}{s} ^ 2$

Here F is the force, d is the distance from the fulcrum, g is the acceleration due to gravity, and M is the mass of the object. I'm using $10 \frac{m}{s} ^ 2$ for the acceleration due to gravity since it turns out this will balance in any gravitational field, we could ignore this term. $\tau$ is the torque.

${\tau}_{\text{right}} = F d = 9 k g \cdot 10 \frac{m}{s} ^ 2 \cdot 4.5 m$
${\tau}_{\text{right}} = 405 \frac{k g \cdot {m}^{2}}{s} ^ 2$

To balance the beam we must exert an equal amount of torque to the left as we have exerted to the right. That extra torque is the difference between the two calculated above.

${\tau}_{\text{x" = tau_"right" - tau_"left}} = 215 \frac{k g \cdot {m}^{2}}{s} ^ 2$

The correct answer is the one which produces a torque of 215 Nm.

For example, if you chose a mass of 21.5 kg, you could balance the system by placing that mass at a distance which would produce 215 Nm of torque.

$\tau = M \cdot g \cdot d$
$215 N m = 21.5 k g \cdot 10 \frac{m}{s} ^ 2 \cdot d$
$1 m = d$

In words: a 21.5 kg mass will balance this beam if placed 1 meter to the left of the fulcrum.