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A uniform wood bar 5 meters long has a mass of 50 kg It is supported at each end by 2 vertical ropes Where must a 130 kg body be placed so that one rope may support twice as much as the other?
Wiki User
∙ 14y ago
Because the bar is uniform the weight is in the center at 2.5 meters.
weight of the bar=9.81m/s2* 50kg= 490.5 newtons.
weight of the body = 130*9.81= 1275.3N.
let's name the supports Ra, Rb
taking the moment around Ra;
sigma Ma=0(anticlockwse)
-490.5N(2.5m)+Rb (5m)=0
Rb =(490.5*2.5)/5=245.25N
sigma F (upwadrs)=0
Ra + Rb =490.5________ Ra=490.5-245.25=245.25N
Now adding the 130kg (1275.3N) mass.
again taking the moment around Ra while the body is at distance x from Ra.
sigma Ma=0 (counterclockwise)
-1275.3(2.5-x)-490.5(2.5)+245.25(5)=0
-3188.25+1275.3x-1226.25+1226.25=0
1275.3x=3188.25
x=3188.25/1275.3
x=2.5m
this means that the body must hang at Ra.
Wiki User
∙ 14y ago
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