A 17.5 kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended fr

Question

A 17.5 kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended from a scale and immersed in water as shown in figure below. The 120. Cm dimension is vertical, and the top of the block is 0.15 cm below the surface of water. What are the magnitude of the forces acting on the top and the bottom of the block due to the surrounding water ? (Use P0=1.0130 times 10 5 N/m2.) F top= F bottom What is the reading of the spring scale ? Your response doffers significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N

Explanation / Answer

Top pressure
= Po + rho x g x h

= 1.01 x 10^5 + 1000 x 9.8 x 0.0515 m

= 101,504 pa

Bottom pressure
= Po + rho x g x h

= 1.01 x 10^5 + 1000 x 9.8 x .175m

= 102,715

Top force
F = P x A
= 101,504 x .01
= 1015.04 N

Bottom force
F = P x A
= 102,715 x .01
= 1027.15 N


Fbuoyant = bottom force - top force

= 12.11 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.