Physics 110 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Griffith, W. Thomas; The physics of everyday phenomena: a conceptual introduction for physics ;4th Edition ISBN 0-07-250977-5
Ch9 CP3. A steel block with a density of 7800 kg/m 3 is suspended from a string in a beaker of water so that the block is completely submerged but not resting on the bottom. The block is a cube with sides of 3 cm (0.03 m).
a.
What is the volume of the block in cubic meters?
b.
What is the mass of the block?
c.
What is the weight of the block?
d.
What is the buoyant force acting on the block?
e.
What tension in the string is needed to hold the block in place?
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Physics 110 Problems - My Solutions
Dr. Hulan E. Jack Jr.
b.
c.
d.
e.
Ch9 CP3. A steel block with a density of 7800 kg/m3is suspended from a string in a beaker of water so that the block is completely submerged but not resting on the bottom. The block is a cube with sides of 3 cm (0.03 m).
a.
What is the volume of the block in cubic meters?
What is the mass of the block?
What is the weight of the block?
What is the buoyant force acting on the block?
What tension in the string is needed to hold the block in place?
AND
e.
a.
Volume of Rectangle = Height x Width x Depth
= Length of Side1 x Length of Side2 x Length of Side 3 =
L
1
*L
2
*L
3
A cube has all three sides equal, so Vol of Cube = L
3
.
b.
mass from density and volume ; density = mass/volume , d =
ρ
= m/Vol, so mass = density x volume m = d* Vol =
ρ
* Vol c.
d.
weight W is the force of gravity on the mass m ; W = mg , where g is the acceleration due to gravity
1.
2.
buoyant force, F buoyant
is the force caused by the difference in fluid pressure pushing up on the bottom of the cube and fluid pressure pushing down on its top.
It is also the weight of the displaced fluid.
3.
3.
By Archimedes Principle the submerged cube displaces a volume of water equal to its own volume Vol.
4.
So, the buoyant force = m displaced water
g = (
ρ water d =
ρ
= m/Vol.
* Vol) g because
The tension, T, in the string -
1.
2.
The string must support the cube.
The cube has two forces acting on it , 1) its weight W pulling it down and 2) the buoyant force pushing up on it.
By Newton’s 1 st Law of Motion
For a body in equilibrium
Σ
F = 0 (Sum of the forces equals zero)
T = Weight - Buoyant Force
W - F buoyant
Page 2 of 3

Physics 110 Problems - My Solutions
Dr. Hulan E. Jack Jr.
b.
c.
d.
e.
Ch9 CP3. A steel block with a density of 7800 kg/m3is suspended from a string in a beaker of water so that the block is completely submerged but not resting on the bottom. The block is a cube with sides of 3 cm (0.03 m).
a.
What is the volume of the block in cubic meters?
What is the mass of the block?
What is the weight of the block?
What is the buoyant force acting on the block?
What tension in the string is needed to hold the block in place?
a.
b.
c.
d.
e.
Volume = L
3
= (0.03m)
3
= (3x10
-2 m)
3
= (3x10
-2 m)* (3x10
-2 m)* (3x10
-2 m); Definition of ( )
3
= (3*3*3) * (10
-2
* 10
-2
* 10
-2
) * (m*m*m) ; 3 of everything
= 27 * 10
-6
m
3
= (2.7*10)*10
-6
m
3
= 2.7 *10
-5
m
3 mass = density of the block * volume
=
ρ block
Vol = 7.8
x 10
= 7.8
x 27 x 10
3
x 10
-6
3
kg/m
(kg/m
3
3
* 27 x 10
)* m
3
-6
m
3
; m
3
cancel , leaving kg
10
3
*10
-6
= (10*10*10)/(10*10*10*10*10*10) = 10
-3
3 10's on top AND 6 10's on the bottom
= 210.6
x 10
-3
kg
= 2.106 x10
2
x 10
-3
kg = 2.106 x 10
-1
kg weight of block = mass g
W= mg = 2.106 x 10 -1 kg x 9.8 m/s 2 = 2.106x9.8 x 10 -1 kg*m/s 2 = 20.64 x 10 -1 N
= 2.064 N kg*m/s 2 = N (Newton)
Buoyant force = weight of displaced water
= mass of water * acceleration due to gravity
= density of water * volume of cube * g
= 1x10 3 kg/m 3 * 27*10 -6 m 3 9.8 m/s 2 = 2.65 *10 -1 N = 0.265 N
Tension in the string = weight of cube - buoyant force = 2.064N - 0.265N = 1.80 N
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