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Standard IX

Physics

Archimedes' Principle

A meter stick...

Question

A meter stick is hung from two spring balances $A$ and $B$ of equal lengths that are located at the $20 c m$ and $70 c m$ marks of the meter stick. Weights of $2.0 N$ are placed at the $10 c m$ and $40 c m$ marks, while a weight of $1.0 N$ is placed at the $90 c m$ mark. The weight of the uniform meter stick is $1.5 N$ . Determine the scale readings of the two balances $A$ and $B$ .

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Solution

Balancing force:
$F_{A} + F_{B} = 2 + 2 + 1.5 + 1$
$F_{A} + F_{B} = 6.5$ ......... $(1)$
Balancing Torque:
ACW Torque: $2 (10) + 2 (40) + F_{B} (20) = 1 (40) + F_{A} (30)$ : Cw Torque
$60 + 20 F_{B} = 30 F_{A}$
$6 + 2 F_{B} = 3 F_{A}$ ....... $(2)$
Solving equations $(1)$ & $(2)$
$F_{A} = 3.8$ N
$F_{B} = 2.7$ N
Readings spring $A \to$ $F_{A} / g = 3.8 / 10 = 0.38$ kg
Spring $B \to F_{B} / g = 2.7 / 10 = 0.27$ kg
(Assuming $g = 10 m / s^{2}$ ).

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Balancing force:FA+FB=2+2+1.5+1FA+FB=6.5 .........(1)Balancing Torque:ACW Torque: 2(10)+2(40)+FB(20)=1(40)+FA(30): Cw Torque60+20FB=30FA6+2FB=3FA .......(2)Solving equations (1) & (2)FA=3.8NFB=2.7NReadings spring A→ FA/g=3.8/10=0.38kgSpring B→FB/g=2.7/10=0.27kg(Assuming g=10m/s2).