A 8.40 kg rock whose density is 4900 kg/m3 is suspended by a string such that half of the rock's volume is under water. 1-What is the tension in the string?
There are 3 forces acting
m*g acting downward
m*g = 8.40*9.8 = 82.32 N
volume of object = mass/density
= 8.40/4900
= 1.714*10^-3 m^3
volume insode water =( 1.714*10^-3)/2 = 8.57*10^-4 m^3
Buoyant force acting upward
FB = density of liquid * volume * g
= 1000 kg/m^3 * 8.57*10^-4 m^3 * 9.8 m/s^2
=8.4 N
Tension is acting upward
balance these 3 forces
T + FB = m*g
T + 8.4 = 82.32
T = 73.9 N
Answer: 73.9 N
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