Question

A meter rule is found to balance at the 48 cm mark. When a body of mass 60 g is suspended at the 6 cm mark, the balance point is found to be at the 30 cm mark. Calculate:

a) The mass of the meter rule.
b) The distance of the balance point from the zero end if the body were moved to the 13 cm mark.

Answer 1

Using the principles of torque balance and static equilibrium, the mass of the meter rule is calculated as 180 g. When the mass is moved to the 13 cm mark, another linear equation can be established and solved to find the new balance point.

Step-by-step explanation:

The question involves concepts of torque balance and center of mass in physics, specifically within the topic of static equilibrium. When the meter rule balances at the 48 cm mark, it means that the rule's center of mass is at that point. When the 60 g mass is suspended at the 6 cm mark and the new balance point is at 30 cm, we can set up a torque balance equation to find the mass of the meter stick (M).

For part a), the initial torque balance is:

(48 cm - 30 cm) * M = (30 cm - 6 cm) * 60 g

Solving this, M = 60*(30-6)/(48-30) = 180 g.

For part b), the new balance equation will be:

(New balance point - 13 cm) * 60 g = (New balance point) * M

Plugging in M = 180 g

(New balance point - 13 cm) * 60 g = (New balance point) * 180 g

(New balance point - 13 cm) = (New balance point) * 180 g : 60 g

(New balance point - 13 cm) = (New balance point) * 3

(New balance point) * 3 - (New balance point) = 13 cm

2 * (New balance point) = 13 cm

(New balance point) = 13 cm : 2

(New balance point) = 6,5 cm

Once solved, we will obtain the required distance from the zero end where the meter stick will balance when the mass is moved to the 13 cm mark.