5.0 kg rock whose density is 4800 kg/m3

5.0 kg rock whose density is 4800 kg/m3

 

 

A 5.0 kg rock whose density is 4800 kg/m3

is suspended by a string such that

half of the rock’s volume is under water.

What is the tension in the string?

 

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from Newton’s first law,

T + F_B – mg = 0

 

> tension, T

> buoyant force, F_B = ρ_w g V

 

T = mg – F_B

 

 

 

rock의 밀도 = 질량 / 부피 이므로,

ρ_r = m / V

 

V = m / ρ_r

 

 

 

T = mg – F_B

= mg – ρ_w g (V / 2)

= g[m – (ρ_w / 2) V]

= g[m – (ρ_w / 2) (m / ρ_r)]

= mg[1 – (ρ_w / 2ρ_r)]

 

 

 

[half of the rock’s volume is under water]

[rock의 부피의 1/2이 물에 잠긴 조건]

T = mg[1 – (ρ_w / 2ρ_r)]

= 5 × 9.8[1 – (1000 / (2×4800))]

= 43.9 N

 

 

 

답: 44 N

 

 

 

 

[키워드] 부력과 장력 기준문서, 장력과 부력 기준문서