Gravitation and the Principle of Superposition Problems and Solutions

 Problem#1

Find the magnitude and direction of the net gravitational force on mass A due to masses B and C in Fig.  1. Each mass is 2.00 kg.

Fig.1

Answer:
From the universal law of gravitation the gravitational force of attraction between the mass is

F = Gm₁m₂/r₁₂²

Then, the magnitude and direction of the net gravitational force on mass A due to masses B and C in Fig.  1a is

F_A = F_AC + F_AB = Gm_Am_C/r_AC² + Gm_Am_B/r_AB² = Gm² [1/r_AC² + 1/r_AB²]
F_A = (6.67 x 10⁻¹¹ Nm²/kg²)(2 kg)²[1/(0.1 m)² + 1/(0.5 m)²]
F_A = 2.77 x 10⁻⁸ N towards mass B

Fig.2

and the magnitude and direction of the net gravitational force on mass A due to masses B and C in Fig.  1b is

F_A = ─ F_AC + F_AB = ─Gm_Am_C/r_AC² + Gm_Am_B/r_AB² = Gm² [1/r_AB² ─ 1/r_AC²]
F_A = (6.67 x 10⁻¹¹ Nm²/kg²)(2 kg)²[1/(0.1 m)² ─ 1/(0.5 m)²]
F_A = 2.56 x 10⁻⁸ N towards mass B

Problem#2
Answer:In Fig. 3, two point particles are fixed on an x axis separated by distance d. Particle A has mass m_A and particle B has mass 3.00m_A. A third particle C, of mass 75.0m_A, is to be placed on the x axis and near particles A and B. In terms of distance d, at what x coordinate should C be placed so that the net gravitational force on particle A from particles B and C is zero?

Fig.3

Answer:
If the distance between mass C and A is x, the distance between mass C and B is d ─ x, so thatMass C must be placed between masses A and B to produce forces at C as shown in figure 4.

F_AC = Gm_Am_C/x² and F_AB = Gm_Bm_A/d²

Fig.4

So that the total gravitational force at mass C due to mass A and B is zero, then

F_AC = F_AB
Gm_Am_C/x² = Gm_Bm_A/d²
75.0m_A/x² = 3.00m_A/d²
5/x = 1/d
x = 5d

So the position of particle C will be at ─5d from point A.

Problem#3
In Fig. 5, three 5.00 kg spheres are located at distances d₁ = 0.300 m and d₂ = 0.400 m.What are the (a) magnitude and (b) direction (relative to the positive direction of the x axis) of the net gravitational  force on sphere B due to spheres A and C?

Fig.5

Answer:
The forces acting on mass B are shown in figure 6.
From the universal law of gravitation the gravitational force of attraction between the mass is

F = Gm₁m₂/r₁₂²
The magnitude of the interaction force between mass B and mass A and mass C is respectively

F_BA = Gm_Am_B/r_AB² = (6.67 x 10⁻¹¹ Nm²/kg²)(5 kg)²/(0.300 m)² = 1.85 x 10⁻⁸ N
F_BC = Gm_Cm_B/r_CB² = (6.67 x 10⁻¹¹ Nm²/kg²)(5 kg)²/(0.400 m)² = 1.04 x 10⁻⁸ N

Fig.6

the magnitude and direction of the net gravitational force on mass B due to masses A and C in Fig. 5 is

F_B² = F_BA² + F_BC²
F_B² = (1.85 x 10⁻⁸ N)² + (1.04 x 10⁻⁸ N)²
F_B = 2.13 x 10⁻⁸ N

The direction of the force relative to the +x─axis is

tan θ = F_BA/F_BC = 1.85 x 10⁻⁸ N/1.04 x 10⁻⁸ N
θ = tan⁻¹ (1.7788) = 60.65⁰

Problem#4
Two uniform spheres, each of mass 0.260 kg, are fixed at points A and B (Fig. 7). Find the magnitude and direction of the initial acceleration of a uniform sphere with mass 0.010 kg if released from rest at point P and acted on only by forces of gravitational attraction of the spheres at A and B.

Fig.7

Answer:
Mass P is attracted to both masses. Since each mass below mass P has the same value the x-components of the gravitational forces cancel. The y-components of each force add to equal the net force. We'll use the gravitation equation for the y-direction and plug it into Newton's second law to determine the acceleration of mass P. First we need to determine the angle between the vertical (6 cm) and distance between mass P and mass A (10 cm.)

cos θ = 6/10

Now we'll set the gravitational equation equal to ma to determine the acceleration of mass 2

∑F = ma
2F cos θ = m_Pa

Fig.8

With F = Gmm_P/r²
(2Gmm_P/r²) cos θ = m_Pa
a = (2Gm/r²) cos θ
a = (6.67 x 10⁻¹¹ Nm²/kg²)(0.260 kg)(0.6)/(0.01 m)²
a = 5.6 x 10⁻⁸ m/s²  

Since the net force is downward the acceleration is downward.