Ans. By placing the weight of 80 gf at a distance 20 cm to the right of point O. 11. Fig. 1.33 shows a uniform metre rule placed on a fulcrum at its mid-point O and having a weight 40 gf at the 10 cm mark and a weight of 20 gf at the 90 cm mark. (i) Is the metre rule in equilibrium ? If not,how will they rule the term ? how can the rule be brought in equilibrium by using an additional weight of 40 g f?
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A uniform metre rule placed on a fulcrum at its mid-point O and having a weight 40 gf at the 10 cm mark and weight of 20 gf at the 90 cm mark.
(i) Is the metre rule in equilibrium? If not. how will the rule turn?
(ii) How can the rule be brought in equilibrium by using an additional weight of 40 gf?
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Solution
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(i) Anticlockwise moment =40gf×(50−10)cm
=40gf×40cm=1600gf×cm
Clockwise moment =20gf×(90−50)=20gf×40cm
=800gf×cm
Anticlockwise moment is not equal to clockwise moment. Hence the metre rule is not inequilibrium and it will turn anticlockwise.
(ii)To balance it,40gf weight should be kept on right hand side so as to produce a clockwisemoment about the middle point. Let its distance from the middle be d cm. Then,clockwise moment =20gf×40cm+40gf×dcm
From the principle of moments,
Anticlockwise moment= Clockwise moment
=40gf×40cm=20gf×40+40×dcm
1600−800=40gf×dcm
So,d=
40gf
800gfcm
=20cm (on the other side)
Hence, by placing the additional weight of 40 gf at the 70 cm mark the rule can be brought in equilibrium