A cook uses a fire tong of length 28 cm to lift a piece of burning coal of mass 250 g. If he applies his effort a distance of 7 cm from the fulcrum, what is the effort in SI unit.

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Solution

Given load = $250 g f$ , load arm $= 28 c m,$
effort arm = $7 c m$ and effort $= E .$
Now, load $\times$ load arm = effort $\times$ effort arm
$250 \times 28 = E \times 27$
$E = \frac{250 \times 28}{7} = 1000 g f = 1 k g f$ or $10 N$

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A cook uses a fire tong of length 28 cm to lift a piece of burning coal of mass 250 g. If he applies his effort a distance of 7 cm from the fulcrum, what is the effort in SI unit.

Given load =250gf, load arm =28cm,effort arm = 7cm and effort =E.Now, load × load arm = effort × effort arm250×28=E×27E=250×287=1000gf=1kgf or 10N

Given load = $250 g f$ , load arm $= 28 c m,$
effort arm = $7 c m$ and effort $= E .$
Now, load $\times$ load arm = effort $\times$ effort arm
$250 \times 28 = E \times 27$
$E = \frac{250 \times 28}{7} = 1000 g f = 1 k g f$ or $10 N$