(c9p30) A 11.89 -kg block of metal is suspended from a scale and immersed in wat

Question

(c9p30) A 11.89 -kg block of metal is suspended from a scale and immersed in water as in Figure P9.30. The dimensions of the block are 14.0 cm x 12.0 cm x 12.0 cm. The 14.0-cm dimension is vertical, and the top of the block is 13.0 cm below the surface of the water. (a) What are the forces exerted by the water on the top and bottom of the block? (Take P0 = 1.013

A 11.89 -kg block of metal is suspended from a scale and immersed in water as in Figure P9.30. The dimensions of the block are 14.0 cm x 12.0 cm x 12.0 cm. The 14.0-cm dimension is vertical, and the top of the block is 13.0 cm below the surface of the water. (a) What are the forces exerted by the water on the top and bottom of the block? (Take P0 = 1.013 x 10^5 N/m^2.) (top force, bottom force) (b) What is the reading of the spring scale? (c) Notice that the buoyant force equals the difference between the forces at the top and bottom of the block. What is the buoyant force?

Explanation / Answer

Force = pressure * area
Pressure at depth 'h' = Po + hdg
So, F top = (1.013 *10^5+ 0.14*1000*9.8) * (0.12*0.12) = 1478.48 N
F bottom = (1.013 *10^5 + 0.27*1000*9.8) * (0.12*0.12) = 1496.82 N

(b) Reading on spring scale = actual weight - (F bottom - F top) = m*g - (1496.82 - 1478.48) = 98.182 N

(c) Difference =(1496.82 - 1478.48) = 18.34 N

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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