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Saturday, December 28, 2019

Chapter 8.4 - Solved Examples on Gravitational Force

In the previous sectionwe saw the magnitude of direction of the gravitational force of attraction . We also saw a solved example. In this section we will see a few more solved examples

Solved example 8.8
Find the unit of the universal gravitational constant G
Solution:
• We have: |FG|=Gm1m2|r|2
• So we get: G=|r|2|FG|m1m2
• Thus the unit is: Nm2kg2

Solved example 8.9
Two heavy bodies of masses 40 kg and 60 kg attract each other with a force of 4 × 10-5 N. If G is 6 × 10-11 N m2 kg-2, find the distance between them
Solution:
• We have: |FG|=Gm1m2|r|2
• Substituting the given values, we get: 4×105(N)=6×1011(Nm2kg2)(40(kg)×60(kg)|r|2)
4×105=(1.44×107(m2)|r|2)
|r|2=(1.44×107(m2)4×105)=0.0036(m2)
|r|=0.06m

Solved example 8.10
Two bodies A and B of masses M and 4M respectively, are placed at a distance of 2.73 m apart. Another body C of mass m is to be placed in between A and B in such a way that, the net force on C is zero. What is the distance of C from A
Solution:
1. Fig.8.14 (a) below shows the 3 bodies in a line
Let C be at a distance of  x m from A
Fig.8.14
2. Fig.b shows the free body diagram of C
(i) We have: FG(A,C)=[GM×m|rA,C|3]rA,C
FG(A,C)=[GM×mx3](x)(ˆi)
FG(A,C)=[GMmx2]ˆi

(ii) We have: FG(B,C)=[Gm×4M|rB,C|3]rB,C
FG(A,C)=[Gm×4M(2.73x)3](2.73x)ˆi
FG(A,C)=[4GmM(2.73x)2]ˆi
3. Net force = FG(A,C)+FG(B,C)
⇒ Net force = [GMmx2]ˆi+([4GmM(2.73x)2]ˆi)
4. But net force must be zero. So we get:
[GMmx2]ˆi=[4GmM(2.73x)2]ˆi
• We have an equality of two vectors. So their magnitudes must be the same. We can write:
[GMmx2]=[4GmM(2.73x)2]
[1x2]=[4(2.73x)2]
(2.73x)2=4x2 
⇒ (2.73-x) = 2x (Taking roots on both sides)
⇒ 2.73 = 3x
⇒ x = 0.91 m

Solved example 8.11
Two bodies A and B have equal masses. The gravitational force of attraction between them is FG. 25% of the mass of A is transferred to B. What is the new force of attraction
Solution:
1. Let the initial masses of the two bodies be m each 
• Then we can write: FG=[Gm2|r|3]r
• Where r is the distance vector between A and B
2. Final mass of A = (m-0.25m) = 0.75m
• Final mass of B = (m+0.25m) = 1.25m
3. New force FG(new)=[G(0.75m)(1.25m)|r|3]r=[0.9375Gm2|r|3]r
4. Dividing (3) by (1), we get:
FG(new)FG=([0.9375Gm2|r|3]r)÷([Gm2|r|3]r)=0.9375
FG(new)=0.9375FG

Solved example 8.12
Two bodies A and B have masses m and M respectively. The gravitational force of attraction between them is F. A third body C is placed in contact with A. The body C has a mass of 2m. What is the force on B due to A? What is the total force on B ?
Solution:
Part (a):
• Initially, there are only two masses A and B
• At that time, the force is given to be F
• We can write:
F=FG(A,B)=[GmM|r|3]r
• Where r is the distance vector between A and B
Part (b):
• When C is placed near A, the force on B due to C is given by:
FG(C,B)=[G×2m×M|r|3]r
• Taking ratios, we get:
FG(C,B)FG(A,B)=FG(C,B)F=([G×2m×M|r|3]r)÷([G×m×M|r|3]r)
FG(C,B)F=2
FG(C,B)=2F
• Thus we can write:
Total force on B FG(A,B)+FG(C,B)=F+2F=3F

Solved example 8.13
In fig.8.15(a) below, three equal masses of m kg each are fixed at three corners A, B and D of the square of side a
Fig.8.15
What is the force acting on a mass 1 kg placed at the fourth corner C ?
Solution:
1. Any square will lie on a plane. So this is a 2D problem. We will need the x and y axes only. There is no need for the z-axis
• Assume that, the origin of the coordinate axes is at A. This is shown in fig.b
2. For this square:
(i) C will be equidistant from B and D
C will be at a distance of (2)a from A
3. Now we can begin the calculations:
• The distance vectors are shown in fig.8.16(a) below:
Fig.8.16
• The free body diagram of the 1 kg mass at C is shown in fig.16(b)
4. Force due to mass at B
(i) FG(B,C)=[Gm×1|rB,C|3]rB,C
FG(B,C)=[Gma3]rB,C
(ii) So our next task is to write rB,C in component form
• To reach C from B:
    ♦ We have no horizontal distance to travel 
    ♦ We travel only a distance BC vertically up
• So the components of rB,C are (0)ˆi and aˆj
• We can write: rB,C=aˆj
(iii) So the result in (i) becomes: FG(B,C)=[Gma3]aˆj
5. Force due to mass at D
(i) FG(D,C)=[Gm×1|rD,C|3]rD,C
FG(D,C)=[Gma3]rD,C
(ii) So our next task is to write rD,C in component form
• To reach C from D:
    ♦ We have no vertical distance to travel 
    ♦ We travel only a distance DC horizontally to the right
• So the components of rD,C are (a)ˆi and 0ˆj
• We can write: rD,C=aˆi
(iii) So the result in (i) becomes: FG(D,C)=[Gma3]aˆi
6. Force due to mass at A:
(i) FG(A,C)=[Gm×1|rA,C|3]rA,C
FG(A,C)=[Gm(2a)3]rA,C 
(ii) So our next task is to write rA,C in component form
• To reach C from A:
    ♦ First we travel a distance AB horizontally to the right
    ♦ Then we travel a distance BC vertically up
• So the components of rA,C are aˆi and aˆj
• We can write: rA,C=aˆi+aˆj
(iii) So the result in (i) becomes:
FG(A,C)=[Gm(2a)3](aˆi+aˆj)
FG(A,C)=([Gm(2a)3]aˆi)+([Gm(2a)3]aˆj)
7. When we add the results in (4), (5) and (6), we get the net force experienced by the mass at C
So we can write:
The net force experienced by the mass at C
FG(B,C)+FG(D,C)+FG(A,C)
([Gma3]aˆj)+([Gma3]aˆi)+([Gm(2a)3]aˆi)+([Gm(2a)3]aˆj)
(i) First we will add the horizontal terms:
([Gma3]aˆi)+([Gm(2a)3]aˆi)
[Gma3](aˆi+[122]aˆi)
[Gma2](1+122)ˆi
(ii) Next we add the vertical terms:
([Gma3]aˆj)+([Gm(2a)3]aˆj)
[Gma3](aˆj+[122]aˆj)
[Gma2](1+122)ˆj
8. We can write:
■ The resultant force on C has two components:
• The horizontal component is: [Gma2](1+122)ˆi
• The vertical component is: [Gma2](1+122)ˆj
9. Note that, the magnitudes of both components are the same
So we can write:
■ The resultant force on C has two components:
• The horizontal component is: pˆi
• The vertical component is: pˆj
Where p=[Gma2](1+122)
This is shown in fig.8.17(a) below:
Fig.8.17
10. We have two vectors
    ♦ The magnitudes of the two vectors are equal
    ♦ The two vectors are perpendicular to each other
• So the magnitude of the resultant vector will be given by:
p2+p2=2p2=2p (Details here)
2[Gma2](1+122)
[Gma2](2+12)
• Also, the resultant will be making an angle of 45o with the horizontal. This is shown in fig.8.17(b)

Another method:
• In (6) we wrote: rA,C=aˆi+aˆj
• We can use trigonometry and write:
rA,C=(acos45)ˆi+(asin45)ˆj
• The reader is advised to write all steps using this alternate method and prove that, the resultant force on the mass at C is the same as that obtained above

In the next section, we will see gravitational force of attraction between extended objects



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