Question

Think about an arrangement of two masses m1 and m2 associated by a massless slender bar, the volumes of the first and second masses are V1 = 840 cm3 and V2 = 330 cm3 correspondingly. The framework drifts in a liquid so that a portion of the primary mass and the whole second mass are beneath the outside of the liquid. Discover the total of the majority if the density of the liquid is ρfluid = 1.23 g/cm3

m, V g 1 m 2, 12

Mtotal = Unit in kg

Locate the each mass if the 2nd mass is five times more than the 1st mass. The tension in the bar, what is it?

m1 = Unit in kg

m2 = Unit in kg

T = Unit in N

Answer 1

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FB = mig+T - FBI EFy= IT hogy Т. ↑ FB₂ Buoyant force. m2 IN magal T & F82 ung Substract Equation @ 40 FB1 Fer mig = mag- FBit FB2 = (mith2g A 1 FBI NEL, F1 = Sound & g x vi 6 = 1.2349.81.X 840 X10 xlovo 10.14 N For a a 1.23 X1600 x 9.81 x 9.81 x 330xió? 3.98 N from Eqreation ④ mit M2= FB + FB2 10-14 +2.98 = 1.569 9.P) Given M2=5 un put in mita5m, = 1.569 so mia m2 = 1.32 kg Now, T = FB - mig g 0.26 kg - 10.14 - 0.26x9.81 = 7:59 N-] CS Scanned with CamScanner