Question

Particle of mass m1 = 0.67 kg is a distance d = 23 cm from one endof a uniform rod with length L = 3.0 m and mass M = 5.0 kg.

What is the magnitude of the gravitational force F on the particlefrom the rod?

Answer 1

Answer:

F = 3 x 10⁻¹⁰ N

Explanation:

given,

mass of particle, m₁ = 0.67 Kg

distance, d = 23 cm = 0.23 m

Length of rod, L = 3 m

mass of rod, M = 5 Kg

Mass of the element of the length dr of the rod is , dm = m dr

mass per unit length M = m/ L

we know,

Force between two particle

$F = \dfrac{GMm_1}{r^2}$

$dF = \dfrac{GMm_1}{L}\dfrac{dr}{r^2}$

integrating both side with limit d to d + L

$dF = \dfrac{GMm_1}{L}\int_{d}^{d+L}\dfrac{dr}{r^2}$

$F = \dfrac{GMm_1}{d(d+L)}$

$F = \dfrac{6.67 \times 10^{-11}\times 0.67\times 5}{0.23(3.23)}$

F = 3 x 10⁻¹⁰ N

The magnitude of gravitational force F is equal to F = 3 x 10⁻¹⁰ N