Archimedes' lever

Classical Mechanics Level 1

Archimedes is trying to balance the Earth on a simple lever.

If Archimedes' distance from the fulcrum is 1 m 1m at what distance from the fulcrum the earth should be placed on the other side so that the lever balances perfectly (i.e. it is not inclined on either side).

If the distance is D m Dm submit − ( log ⁡ ( D ) 2 + log ⁡ 3 ) -\left(\dfrac{\log\left(D\right)}{2}+\log3\right)

Assumptions:

  • M a s s o f E a r t h = 6.3 × 1 0 24 K g Mass\space of\space Earth=6.3\times 10^{24}Kg

  • M a s s o f A r c h i m e d e s = 70 K g Mass\space of\space Archimedes = 70Kg


The answer is 11.

Zakir Husain by Zakir Husain , 41, India

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1 solution

Tom Engelsman
Apr 17, 2021

In order for Archimedes and the Earth to balance the lever, their torques about the fulcrum must be equal to each other. This can be expressed as:

T A = T E ⇒ m A g x A = m E g x E ⇒ x E = m A x A m E = 70 ⋅ 1 6.3 × 1 0 2 4 = 1.11 × 1 0 − 21 T_{A} = T_{E} \Rightarrow m_{A}gx_{A} = m_{E}gx_{E} \Rightarrow x_{E} = \frac{m_{A}x_{A}}{m_{E}} = \frac{70 \cdot 1}{6.3 \times 10^24} = 1.11 \times 10^{-21} meters.

Hence, − log ⁡ ( 3 ) − log ⁡ ( − 1.11 × 1 0 − 21 ) 2 = 11 . -\log(3) - \frac{\log (-1.11 \times 10^{-21})}{2} = \boxed{11}.

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