Question

A mass of 10 kg hangs from one end of a 1 m long light rod that is pivoted 0.3 m from that end.
(a) what force must be applied at the 0.6 m mark to balance the rod?
(b) if force of 20N is hung from the 0.5 m mark what force must be hung from the 1 m mark to balance the rod?

Answer 1

Answer on Question #64417-Physics-Mechanics-Relativity

A mass of $m = 10$ kg hangs from one end of a $l = 1$ m long light rod that is pivoted $d = 0.3$ m from that end.

(a) What force must be applied at the $s = 0.6$ m mark to balance the rod?

(b) If force of $F_{1} = 20N$ is hung from the $d_{1} = 0.5$ m mark what force must be hung from the $l = 1$ m mark to balance the rod?

Solution

(a) Taking moments about the pivot:

$$Wd = F(s - d)$$ $$F = W \frac{d}{(s - d)} = mg \frac{d}{(s - d)} = 10 \cdot 10 \frac{0.3}{(0.6 - 0.3)} = 100 \, N.$$

(b) Taking moments about the pivot:

$$Wd = F_{1}(d_{1} - d) + F(l - d)$$ $$F = \frac{Wd - F_{1}(d_{1} - d)}{l - d} = \frac{10 \cdot 10 \cdot 0.3 - 20(0.5 - 0.3)}{1 - 0.3} = 37 \, N.$$

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