C:\\Users\ ikki.deering\\Downloads\\FullSizeRender (1).jpg The meterstick shown

Question

C:Users ikki.deeringDownloadsFullSizeRender (1).jpg

The meterstick shown to the right is 100 cm long. It is free to pivot around its center of gravity (CG), which is at the 50-cm mark. There is a 21.0-N block hanging from the 80-cm mark. Decide where each of the other blocks should be placed, one at a time, to balance out the 21.0-N block. A) At what mark on the meter stick would you place a 16.0-N block to balance the 21.0-N block?

B) At what mark on the meter stick would you place a 40.0-N block to balance the 21.0-N block?

Explanation / Answer

A) balancing moment (r x F) about centre of gravity point.

mass must be placed on the other side than 21 N block.

21 x (80 - 50) = 16 x d

d = 39.375 cm from 50 cm mark

Mark =50 - 39.375 = 10.625 cm mark

b ) now balancing moment for 40 N block,

21 x (80 - 50) = 40 x d

d = 15.75 cm

mark = 50 - 15.75 =34.25 cm mark

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