A uniform metre scale is in equilibrium as shown in the diagram below :
(a) Calculate the weight of the metre scale.
(b) Which of the following options is correct to keep the ruler in equilibrium when 40 gf wt is shifted 0 cm mark ? F is shifted towards 0 cm or F is shifted towards 100 cm.
(a) Given,
Fulcrum (F) is at 30 cm mark.
A 40 gf weight is hanging at 5 cm mark.
Let the weight of the metre scale be W gf, acting at its centre of gravity which is at it's midpoint i.e., at 50 cm mark.
Moment of force due to 40 gf weight:
Distance from F = 30 – 5 = 25 cm (anticlockwise)
So, Anticlockwise moment of force = 40 × 25 = 1000 gf cm
Moment due to the weight of the scale :
Weight W gf acts at 50 cm mark.
Distance from F = 50 – 30 = 20 cm (clockwise)
So, Clockwise moment of force = W × 20 gf cm
Using principle of moments:
Clockwise moments of force about the fulcrum = Anticlockwise moments of force about the fulcrum
45 x 25 = W x 20
⇒ 1000 = 20 W
⇒ W = = 50 gf
∴ Weight of the metre scale = 50 gf
(b) Now the 40 gf weight is shifted to the 0 cm mark.
New moment of force due to 40 gf at 0 cm :
Distance from F = 30 – 0 = 30 cm (anticlockwise)
Anticlockwise moment of force = 40 × 30 = 1200 gf cm
Moment of force due to the metre scale (still acting at its centre of gravity i.e., at 50 cm mark) :
Distance from F = 50 – 30 = 20 cm (clockwise)
Clockwise moment of force = 50 × 20 = 1000 gf cm
Here,
Anticlockwise moment of force > Clockwise moment of force, so the scale will tilt anticlockwise unless we correct this.
To balance, the fulcrum F must be shifted towards 0 cm mark to reduce the arm length of the 40 gf weight.
∴ F is shifted towards 0 cm mark.