A 14.5 kg block of metal measuring 12 cm * 10 cm * 10 cm is suspended from a sca

Question

A 14.5 kg block of metal measuring 12 cm * 10 cm * 10 cm is suspended from a scale and immersed in water as shown in Figure P15.21. The 12.0 cm dimension is vertical and the top of the block is 5.45 cm below the surface of the water.


Figure P15.21*

* I cannot attach the figure because my computer will not me. However, I can describe the figure. There is a block that is immersed in a water cup. It is not floating on the surface of the water cup nor did it sink in the water cup. There is a line attached to the block though.

(a) What are the forces acting on the top and on the bottom of the block? (Use P0 = 1.0130 * 105 N/m2.)

Ftop =

Answer in Newtons

Fbottom =

Answer in Newtons

(b) What is the reading of the spring scale?

Answer in Newtons

c**) Find the difference between the forces on the bottom and the top of the block.

** I can find the difference once I find the top and the bottom of the block.

Explanation / Answer

Force = pressure * area Pressure at depth 'h' = Po + hdg So, F top = (1.013 + 0.0525*1000*9.8) * (0.1*0.1) = 1018.145 N F bottom = (1.013 + 0.1725*1000*9.8) * (0.1*0.1) = 1029.905 N (b) Reading on spring scale = actual weight - (F bottom - F top) = 130.34 N (c) Difference = 1029.905 - 1018.145 = 11.76 N

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