The load arm and the effort arm of a lever are $32 c m$ and $48 c m$ , respectively. How much load can the lever overcome with an effort of $360 k g f ?$

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Solution

Step 1: Given Data
The load's arm $= 32 c m$
Effort arm $= 48 c m$
Applied effort $= 360 k g f$
Let the load which can be overcome with the applied effort $= L k g f$

Step 2: Formula
According to the principle of a lever, we know

$L o a d \times L o a d a r m = E f f o r t \times E f f o r t a r m \Rightarrow \frac{L o a d}{E f f o r t} = \frac{E f f o r t a r m}{L o a d a r m} - - - - E q u a t i o n (I)$

Step 3: Substituting the value of effort arm, load arm, and effort we get-

$\frac{L}{360 k g f} = \frac{48 c m}{32 c m} L = \frac{48 c m \times 360 k g f}{32 c m} L = 540 k g f$
Hence, with the applied effort of $360 k g f$ we can overcome a load of $540 k g f .$

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