Question

A 200 cm meter rule is pivoted at the middle point (at 50 cm point). If the weight of 10 N is hanged from the 30 cm mark and a weight of 20 N is hanged from its 60 cm mark, identify whether the meter rule will remain balanced over its pivot or not.

Answer 1

Answer:

2 Nm

Explanation:

According to the principle of moments, when an object is in rotational equilibrium, then

Total anticlockwise moments = Total clockwise moments

Total anticlockwise moments:

Length of lever arm = (50 – 30) = 20 cm

= 0.20 m

Since the length of the lever arm is the distance from its mid-point, where its balanced force

applied = 10 N

Anticlockwise moment = Lever arm x Force applied

= 0.20 x 10 = 2 Nm

Clockwise moment: Length of lever arm = (60 – 50)

= 10 cm

= 0.10 m.

Since the length of the lever arm is the distance from the mid-point, about which balanced Force

applied = 20 N

Clockwise moment = lever arm x force applied

                             = 0.10 × 20 = 2 Nm

Therefore,

Since the total anti-clockwise moment = total clockwise moment = 2 Nm, according to the  principle of moments, it is in rotational equilibrium ie, the meter rule remains balanced about its  pivot.

Answer 2

Answer:

The meter rule will remain balanced over its pivot.

Explanation:

Using the concept of lever with a fulcrum, the ruler is acting as a lever (rigid rod) that is rotating at a point or fulcrum.

The length of ruler = 1 meter = 100 cm

Point of fulcrum = 50 cm

Load Arm, LA is the distance between load and fulcrum.

Load at 30 cm = 10 N

Load arm 1, LA1 = 50 - 30 = 20 cm = 0.2 m

Load at 60 cm = 20 N

Load arm 2, LA2 = 60 - 50 = 10 cm = 0.1 m

For the ruler to be balanced, the torque due to load 1 and load 2 must be balanced, that is equal in magnitude and opposite in direction.

In the equilibrium condition, the torque provided by the both loads are equal in magnitude (since the friction is neglected) and opposite in direction.

Using the net zero torque condition under equilibrium, the torque applied by the loads is calculated as follows:

                     Torque = Force applied * Lever arm

   The torque due to load 1, = Load 1 * Load Arm 1 = 10 * 0.2 = 2 Nm

   The torque due to load 2, = Load 2 * Load Arm 2 = 20 * 0.1 = 2 Nm

The torque offered by both the loads are equal in magnitude.

Since the two loads are on either side of the fulcrum, the torque directions are opposite to each other.

So, it can be concluded that the meter rule will remain balanced over its pivot.

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