Question

(2) Put 5kg mass at left side (at 2m). This is fixed throughout the experiment! (3) Try to balance by putting 5kg mass at the right side. Write the position of the 5kg mass. Calculate the net torque. Torque= mass * g*length, and the unit for torque is N.m (4) Try to balance by putting 10kg mass at the right side. Write the position of the 10kg mass. Calculate the net torque. (5) Try to balance by putting 20kg mass at the right side. Write the position of the 20kg mass. Calculate the net torque. (6) What do you conclude?

Answer 1

We can conclude that as the mass on the right increases, the distance of the mass towards the right decreases. Also when the two masses balance, the net torque is zero.

What is torque

The torque experienced by an object a given position is the product of the applied force and the perpendicular distance of the object.

When 5 kg mass is at 2 m on the left, another 5 kg at 2 m on the right will balance it.

$\tau _{net} = (2 \times 5 \times 9.8) - (2 \times 5 \times 9.8)\\\\\tau _{net} = 0$

Position of 10 kg mass on the right

Apply principle of moment

$F_1r_1 = F_2r_2\\\\(m_1gr_1) = (m_2gr_2)\\\\r_2 = \frac{m_1gr_1}{m_2g} \\\\r_2 = \frac{m_1 r_1}{m_2} \\\\r _2 = \frac{5 \times 2}{10} \\\\r_2 = 1 \ m$

Net torque

$\tau_{et} = m_2gr_2 - m_1gr_1\\\\\tau_{et} = (10 \times 9.8 \times 1) - (5 \times 9.8 \times 2)\\\\\tau_{et} = 0$

Position of the 20 kg mass

$r_2 = \frac{5 \times 2}{20} \\\\r_2 = 0.5 \ m$

Net torque

$\tau_{et} = m_2gr_2 - m_1gr_1\\\\\tau_{et} = (20 \times 9.8 \times 0.5) - (5 \times 9.8 \times 2)\\\\\tau_{et} = 0$

Thus, we can conclude that as the mass on the right increases, the distance of the mass towards the right decreases. Also when the two masses balance, the net torque is zero.

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