Answer: The mass of the metre stick is 0.33kg
Explanation:
Since the mass of the rock that is suspended from the tip of the meter stick at the 0cm mark is 1kg, we will first need to calculate it's weight.
Weight (force) = mass × acceleration due to gravity (which is approximately 10N/m^2.
Therefore the weight being exerted on that mark on the meter stick by the rock is:
1kg × 10N/m^2 = 10Newton.
If the stick balances like a seesaw when the fulcrum is at the 25cm mark, it means that the distance of the rock from the fulcrum from the point where the rock was suspended (the 0cm mark) is 25cm. This left hand side of the fulcrum is the anticlockwise moment region.
Therefore the anticlockwise moment is 0.25m (convert cm to m) × 10N = 2.5Nm.
The metre stick should be 100cm long which means that the distance of the other end/tip of the stick from the fulcrum is 75cm (100cm - 25cm). Since we don't know the weight or the force being exerted by the stick at this end to keep it balanced like a seesaw, we can use "x" to represent the stick's weight as we solve. Meanwhile, the right hand side of the fulcrum is the clockwise moment region.
Therefore the clockwise moment= 0.75m (convert cm to m)× X = 0.75x
Principle of moments states that the sum of anticlockwise moments = sum of clockwise moments.
Therefore 2.5Nm = 0.75x Nm
x = 2.5/0.75 = 3.33N.
The weight of the metre stick = 3.33N.
Since weight or force = mass × acceleration due to gravity.
i.e 3.33N = mass × 10N/m^2
mass = 3.33/10
mass = 0.33kg.
Therefore the mass of the metre stick is 0.33kg or 330g.
