Question

Three point particles are fixed in position in an xy-plane. Two of them, particle A of mass 5 g and particle B of mass 11 g, are shown in the figure with a separation of d_AB = 0.566 m at angle θ = 30°. Particle C (mass 7 g) is not shown.

The net gravitational force acting on particle A due to B and C is 2.80 × 10⁻¹⁴ N at an angle of −163.8° from the positive x-axis.

What are:

• (a) The x-coordinate of particle C?
• (b) The y-coordinate of particle C?

Answer and Detailed Explanation

Step 1: Place Known Points

Let A = (0, 0).
Using the given angle and distance:
B = (−0.566 × cos(30°), 0.566 × sin(30°)) = (−0.490, 0.283)

Step 2: Compute Force Between A and B

Using Newton’s Law of Gravitation:
F_AB = G × (m_A × m_B) / d²
= (6.674 × 10⁻¹¹) × (5×10⁻³ × 11×10⁻³) / (0.566²)
F_AB = 1.15 × 10⁻¹⁴ N

Step 3: Resolve F_AB into Components

F_AB,x = F_AB × cos(150°) = −9.96 × 10⁻¹⁵ N
F_AB,y = F_AB × sin(150°) = 5.75 × 10⁻¹⁵ N

Step 4: Resolve Net Force into Components

Given:
F_net = 2.80 × 10⁻¹⁴ N at −163.8°
F_net,x = −2.677 × 10⁻¹⁴ N
F_net,y = 8.18 × 10⁻¹⁵ N

Step 5: Calculate Force from C on A

F_AC,x = F_net,x − F_AB,x = −1.681 × 10⁻¹⁴
F_AC,y = F_net,y − F_AB,y = 2.43 × 10⁻¹⁵

Step 6: Compute Magnitude and Direction of F_AC

F_AC = √[(−1.681e−14)² + (2.43e−15)²] ≈ 1.699 × 10⁻¹⁴ N

Step 7: Use Newton’s Law to Find r_AC

r² = G × m_A × m_C / F_AC
r = √(0.1375) = 0.3708 m

Step 8: Determine Direction to C and Find Coordinates

θ = tan⁻¹(2.43e−15 / −1.681e−14) = −8.26°
Since it’s in the second quadrant: θ_actual = 180 − 8.26 = 171.74°

x_C = r × cos(171.74°) = −0.3666 ≈ −0.367 m
y_C = r × sin(171.74°) = 0.0537 ≈ 0.054 m

Final Answers:
(a) x-coordinate of particle C: −0.367 m
(b) y-coordinate of particle C: 0.054 m