A metal alloy rod is submerged 24 cm below the surface of a fresh water pool by
ID: 1525114 • Letter: A
Question
A metal alloy rod is submerged 24 cm below the surface of a fresh water pool by steel cables tied 10cm from each end. It has a length of 80 cm, a mass of 1.5 kg and a uniform square cross sectional area of 6 cm2. Because its density is not uniform its center of mass is located 34 from the left end.
1)What is the force of tension in the left cable?
FLEFT = ?
2)What is the force of tension in the right cable?
FRIGHT = ?
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Do not copy from other people's work. I would like 2 answers here. Thank you.
Explanation / Answer
from the given data
volume of rod, V = A*L
= 6*80
= 480 cm^3
= 4.8*10^-4 m^3
m = 1.5 kg
let F_Left and F_right are the tension in the left and right cables.
In the equilibrium,
Apply, Fnety = 0
F_Left + F_right + Buoyancy - M*g = 0
F_Left + F_right = M*g - Buoyancy
= M*g - rho_water*V*g
F_Left + F_right = 1.5*9.8 - 1000*4.8*10^-4*9.8
F_Left + F_right = 10 N --------(1)
Apply net torque about center of mass.
F_left*(34 - 10) - F_right*(46 - 10) = 0
F_left*24 - F_right*36 = 0 -----(2)
on solving equations 1 and 2
we get
1) F_left = 6 N
2) F_right = 4 N