Question

a weightless rod is acted upon by parallel force of 4 N and 2 N at the ends .length of rod is 3 m . to keep rod in equilibrium 6 N should be applied where

Answer 1

Answer on Question#56965 - Physics - Mechanics - Relativity

A weightless rod is acted upon by parallel force of $F_{1} = 4 \mathrm{~N}$ and $F_{2} = 2 \mathrm{~N}$ at the ends. Length of rod is $L = 3 \mathrm{~m}$ . To keep rod in equilibrium $F = 6 \mathrm{~N}$ should be applied where?

Solution:

According to the principle of moments

$$F _ {2} (L - x) = F _ {1} x$$

Thus

$$x = \frac {F _ {2}}{F _ {1} + F _ {2}} L = \frac {2 \mathrm {N}}{4 \mathrm {N} + 2 \mathrm {N}} 3 \mathrm {m} = 1 \mathrm {m}$$

Answer: $1\mathrm{m}$ from the end where the force $F_{1}$ is applied.

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