A nut has been tightened by a force of 200 N using a 10 cm long spanner. What length of a spanner is required to loosen the same nut with 150 N force? Ans. (13.3 cm)


Given:  

Magnitude of  first Force  = F₁ =   200 N
Length of the first spanner = first moment arm = L₁ =   10 cm = 0.1 m 
Magnitude of  second  Force  = F₂ =   150 N


To Find:

Length of the second spanner = second-moment arm = L₂ = ?

Solution:   

According to the second condition of the equilibrium 

Σ T= 0 

or 

Clockwise Torque = Anticlockwise Torque

F₂ ï½˜ L₂ = F₁ ï½˜ L₁

or

L₂ = `\frac {F₁ ｘ L₁}{F₂ }`
by putting values

L₂ = `\frac {200 N ｘ 0.1 m}{150 N }`

L₂ = 0.133 m

or 

L₂ = 13.3 cm

thus the length of the seconds spanner will be 13.3 cm 

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