Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 14, Problem 27P

A 10.0-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 P14.11b. The 12.0-cm dimension is vertical, and the top of the block is 5.00 cm below the surface of the water. (a) What are the magnitudes of the forces acting on the top and on the bottom of the block due to the surrounding water? (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.

Figure P14.11 Problems 11 and 12.

Chapter 14, Problem 27P, A 10.0-kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended from a scale and

(a)

Expert Solution
Check Mark
To determine

The magnitudes of the forces acting on the top and on the bottom of the block due to the surrounding water.

Answer to Problem 27P

The magnitudes of the forces acting on the top and on the bottom of the block due to the surrounding water is 1018.15N and 1029.91N respectively.

Explanation of Solution

Section 1:

To determine: The magnitudes of the force acting on the top of the block due to the surrounding water.

Answer: The magnitudes of the force acting on the top of the block due to the surrounding water is 1018.15N .

Given information:

The mass of the block is 10.0kg , the height of the block is 12.0cm , the length of the block is 10.0cm , the breadth of the block is 10.0cm and the top of the block is 5.0cm below the surface of the water.

Formula to calculate the absolute pressure at the top of the block is,

Ptop=P0+ρwghtop

Ptop is the absolute pressure at the top of the block.

ρw is the density of water.

g is the acceleration due to gravity.

htop is the depth of the top of the block in the water.

Formula to calculate the force acting on the top of the block is,

Ftop=PtopA=Ptop×(b×l)

Ftop is the force acting on the top of the block.

A is the area of the top face of the block.

l is the length of the block.

b is the breadth of the block.

Substitute P0+ρwghtop for Ptop in the above equation.

Ftop=(P0+ρwghtop)(l×b)

Substitute 1.01325×105Pa for P0 , 1000kg/m3 for ρw , 9.8m/s2 for g , 5.0cm for htop , 10.0cm for l and 10.0cm for b to find Ftop .

Ftop=(1.01325×105Pa+1000kg/m3×9.8m/s2×(5.0cm×102m1cm))×(10.0cm×102m1cm×10.0cm×102m1cm)=101815Pa×0.01m2=1018.15N

Conclusion:

Therefore, the magnitudes of the force acting on the top of the block due to the surrounding water is 1018.15N .

Section 2:

To determine: The magnitudes of the force acting on the bottom of the block due to the surrounding water.

Answer: The magnitudes of the force acting on the bottom of the block due to the surrounding water is 1029.91N .

Given information:

The mass of the block is 10.0kg , the height of the block is 12.0cm , the length of the block is 10.0cm , the breadth of the block is 10.0cm and the top of the block is 5.0cm below the surface of the water.

Formula to calculate the absolute pressure at the bottom of the block is,

Pbottom=P0+ρwghbottom

Pbottom is the absolute pressure at the bottom of the block.

hbottom is the depth of the bottom of the block in the water.

Formula to calculate the force acting on the bottom of the block is,

Fbottom=PbottomA=Pbottom×(b×l)

Fbottom is the force acting on the bottom of the block.

A is the area of the bottom face of the block.

Substitute P0+ρwghbottom for Pbottom in the above equation.

Fbottom=(P0+ρwghbottom)(l×b)

Substitute 1.01325×105Pa for P0 , 1000kg/m3 for ρw , 9.8m/s2 for g , (5.0cm+12.0cm) for hbottom , 10.0cm for l and 10.0cm for b to find Fbottom .

Ftop=(1.01325×105Pa+1000kg/m3×9.8m/s2×((5.0cm+12.0cm)×102m1cm))×(10.0cm×102m1cm×10.0cm×102m1cm)=102991Pa×0.01m2=1029.91N

Conclusion:

Therefore, the magnitudes of the force acting on the bottom of the block due to the surrounding water is 1029.91N .

(b)

Expert Solution
Check Mark
To determine

The reading of the spring scale.

Answer to Problem 27P

The reading in the spring scale is 86.24N .

Explanation of Solution

Given information:

The mass of the block is 10.0kg , the height of the block is 12.0cm , the length of the block is 10.0cm , the breadth of the block is 10.0cm and the top of the block is 5.0cm below the surface of the water.

The scale reading when the block is immersed in water is equal to the tension in the chord supporting the block.

Apply the equilibrium condition.

Fy=0T+FbottomFtopmg=0

T is the tension in the chord.

m is the mass of the block.

Rearrange the above equation for T .

T=Ftop+mgFbottom

Substitute 1029.91N for Fbottom , 10.0kg for m , 9.8m/s2 for g and 1018.15N for Ftop to find T .

T=1018.15N+10.0kg×9.8m/s21029.91N=86.24N

Conclusion:

Therefore, the reading in the spring scale is 86.24N .

(c)

Expert Solution
Check Mark
To determine

To show: That the buoyant force equals the difference between the force at the top and bottom of the block.

Explanation of Solution

Given information:

The mass of the block is 10.0kg , the height of the block is 12.0cm , the length of the block is 10.0cm , the breadth of the block is 10.0cm and the top of the block is 5.0cm below the surface of the water.

When an object is immersed in a liquid, then the upward force exerted on it by fluid is called buoyant force and its magnitude equals the weight of the liquid displaced by the object.

Formula to calculate the buoyant force is,

B=ρwVg

B is the buoyant force.

ρw is the density of water.

V is the volume of block.

g is the acceleration due to gravity.

Substitute 1000kg/m3 for ρw , (12.0cm×10.0cm×10.0cm) for V and 9.8m/s2 for g to find B .

B=1000kg/m3×((12.0cm×10.0cm×10.0cm)×106m31cm3)×9.8m/s2=11.76N

From part (a), the difference between the force at the top and bottom of the block is,

FbottomFtop=1029.91N1018.15N=11.76N

From the above calculations it is clear that,

B=FbottomFtop

Conclusion:

Therefore, the buoyant force equals the difference between the force at the top and bottom of the block.

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Chapter 14 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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