A nut can be opened by a lever of length $0.25 m$ by applying a force of $80 N$ . What should be the length of the lever if a force of $32 N$ is enough to open the nut?

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Solution

Step 1: Given data,
Load $= 80 N,$
Load arm $= 0.25 m,$
Effort $= 32 N .$
Formula:
Principle of the lever.
$L o a d \times L o a d a r m = E f f o r t \times E f f o r t a r m$
Step 2: Calculate the length of the effort arm
Substitute the values in the above formula.
Effort arm = $\frac{L o a d \times L o a d a r m}{E f f o r t}$
$E f f o r t a r m = \frac{80 N \times 0.25 m}{32 N} = \frac{20}{32} = 0.625 m$
Hence, the length of the lever is $0.625 m .$

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A nut can be opened by a lever of length 0.25m by applying a force of 80N. What should be the length of the lever if a force of 32N is enough to open the nut?

A nut can be opened by a lever of length $0.25 m$ by applying a force of $80 N$ . What should be the length of the lever if a force of $32 N$ is enough to open the nut?