Asked by Obert
A uniform metres rod of weight 100N carries a weight of 40N and 60N suspended from the 20Cm and 90cm marks respectively, where will your pivot a knife edge to balance the rod.
Answers
Answered by
bobpursley
vertical forces:
100+40+60-KE=0 or KE=200N
now sum moments around anywhere, I feel frisky, and will do it from the 20cm point: d will be the distance from the 20 cm point.
-KE*d+100*30+60*70=0
d=(3000+4200)/200= 72/2=36cm from the 20cm mark, or at the 56cm mark on the stick.
check my math.
100+40+60-KE=0 or KE=200N
now sum moments around anywhere, I feel frisky, and will do it from the 20cm point: d will be the distance from the 20 cm point.
-KE*d+100*30+60*70=0
d=(3000+4200)/200= 72/2=36cm from the 20cm mark, or at the 56cm mark on the stick.
check my math.
Answered by
Nakul
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