Figure 3 shows a uniform metre rule pivoted at the 30 cm mark. It is balanced by a weight of 2 N suspended at the 5 cm mark.

      

Figure 3 shows a uniform metre rule pivoted at the 30 cm mark. It is balanced by a weight of 2 N suspended at the 5 cm mark.
22metrerule4192017831.jpg
Determine the weight of the metre rule

  

Answers


John
W x 0.2 = 2 x 0.25 = 0.25

W = $\frac{2 \times 0.25}{0.2}$ = 2.5 N
johnmulu answered the question on April 19, 2017 at 07:15


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