Question

1. A meter rule is found to balance at the 48cm mark. When a body of mass 60g is suspended at the 6cm mark the balance point is found to be at the 30cm mark. Calculate, i) the mass of the meter rule 11) the distance of the balance point from the zero end, if the body were moved to the 13cm mark. ​

Answer 1

Hi there!

We can use a summation of torques to solve.

Recall the equation for torque:
$\large\boxed{\Sigma \tau = rF}$

r = distance from fulcrum (balance point)

F = force (in this instance, weight, N)

We can set the fulcrum to be the balance point of 30 cm.

Thus:
Meter ruler:

Center of mass at 48 cm ⇒ 48 - 30 = 18 cm

Object:
At 6cm ⇒ 30 - 6 = 24 cm

For the ruler to be balanced:
$\large\boxed{\Sigma \tau_{cc} = \Sigma \tau_{ccw}}$

Thus:
$M_Rg(18) = 60g(24)\\M_R = \frac{60(24)}{18} = \boxed{80 g}$

The mass of the ruler is 80 grams.

If the body were moved to 13 cm:
B (balance point) - 13 = distance of object

48 - B = distance from ruler center of mass to balance point

$80g(48 - B) = 60g(B - 13)\\\\3840 - 80B = 60B - 780\\\\4620 = 140B\\\boxed{B = 33 cm}$

The new balance point would be 33cm from the zero end.