Question

a uniform scale of weight gf is balanced at 30cm mark when the weight of 80gf and 60gf acts at 5cm and 45cm mark respectively. what force must be applied at 20cm to balance the meter scale ??


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Answer 1

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Let the meter scales is balanced at X cm mark on the scale.

At the balancing condition, Anticlock wise moment must be equal to clock wise moment.

Clockwise moment, about the balancing point is

= moment by 80 gf

=80gf×(80−X)cm

=6400gfcm−80Xgfcm.....(1)

Anticlock wise moment about the balancing point is

= moment by 10 gf + moment by 50 gf

=10gf×(X−10)+50gf×(X−50)

=10Xgfc,−100gfcm+50Xgfcm−2500gfcm.....(2)

At the balancing condition, Clock wise at equilibrium = Anti clock wise moments.

6400−80X=100X−100+50X−2500

6400+2600=140X⇒140X=8000

⇒X=1408000=64.28

Scale is balanced at 64.28 cm from the beginning.

Answer 2

||✪✪ CORRECT QUESTION ✪✪||

A uniform scale of weight 50gf is balanced at 30cm mark when the weight of 80gf and 60gf acts at 5cm and 45cm mark respectively. Where must a weight of 80gf be suspended to balance the metre scale? ??

|| ✰✰ ANSWER ✰✰ ||

❁❁ Refer To Image For Diagram.. ❁❁

As we can see now, 50 gm is located at 10 cm and 100 gm is at 45 cm distance.

Now using Principal of Moments in Equilibirium :-

==>> F1 * D1 = F2 * D2

Putting values ,

→ (15 * 50) + (50*10) = 100 * x

→ 750 + 500 = 100x

→ 100x = 1250

→ x = 12.5cm .

Now,

→ Since, Distance = 12.5cm from middle of the scale.

So,

→ 80 gm is located at = (60+12.5) = 72.5cm on the Meter Scale.