Question

Two uniform spheres, each of mass 0.260 kg are fixed at points 'A' and 'B' as shown in the figure. Find the magnitude and direction of the initial acceleration of a sphere with mass 0.010 kg if it is released from rest at point 'P' and acted only by forces of gravitational attraction of sphere at 'A' and 'B' (give your answer in terms of G).

Answer 1

Magnitude of acceleration: $4.16\sqrt{5}G \text{ m/s}^2$

Direction of acceleration: Vertically downwards

Answer 2

The problem asks for the initial acceleration of a small sphere released from rest at point P, subjected to gravitational forces from two larger spheres at points A and B.

1. Define the Coordinate System and Given Values: Let the origin be at the midpoint of AB.

• Point A: $(-0.100 \text{ m}, 0)$
• Point B: $(0.100 \text{ m}, 0)$
• Point P: $(0, 0.200 \text{ m})$
• Mass of spheres at A and B: $M = 0.260 \text{ kg}$
• Mass of sphere at P: $m = 0.010 \text{ kg}$

2. Calculate Distances from P to A and P to B: The distance $r_{AP}$ between P$(0, 0.200)$ and A$(-0.100, 0)$ is:

$r_{AP} = \sqrt{(-0.100 - 0)^2 + (0 - 0.200)^2} = \sqrt{(-0.100)^2 + (-0.200)^2}$

$r_{AP} = \sqrt{0.0100 + 0.0400} = \sqrt{0.0500} \text{ m}$

The distance $r_{BP}$ between P$(0, 0.200)$ and B$(0.100, 0)$ is:

$r_{BP} = \sqrt{(0.100 - 0)^2 + (0 - 0.200)^2} = \sqrt{(0.100)^2 + (-0.200)^2}$

$r_{BP} = \sqrt{0.0100 + 0.0400} = \sqrt{0.0500} \text{ m}$

Since $r_{AP} = r_{BP}$, let $r = \sqrt{0.0500} \text{ m}$. Thus, $r^2 = 0.0500 \text{ m}^2$.

3. Calculate the Magnitudes of Gravitational Forces: According to Newton's Law of Universal Gravitation, $F = \frac{GMm}{r^2}$.

• Force due to sphere A on sphere P ($F_A$):

$F_A = \frac{G \times M \times m}{r_{AP}^2} = \frac{G \times 0.260 \text{ kg} \times 0.010 \text{ kg}}{0.0500 \text{ m}^2}$

$F_A = \frac{0.0026 G}{0.05} = 0.052 G \text{ N}$

• Force due to sphere B on sphere P ($F_B$):

$F_B = \frac{G \times M \times m}{r_{BP}^2} = \frac{G \times 0.260 \text{ kg} \times 0.010 \text{ kg}}{0.0500 \text{ m}^2}$

$F_B = \frac{0.0026 G}{0.05} = 0.052 G \text{ N}$

So, $F_A = F_B = 0.052 G$.

4. Determine the Direction and Components of Forces: The forces are attractive, acting along the lines joining the centers of the spheres. Let's find the components of these forces. The vector from P to A is $\vec{r}_{PA} = A - P = (-0.100, -0.200)$. The vector from P to B is $\vec{r}_{PB} = B - P = (0.100, -0.200)$.

The force $\vec{F_A}$ acts in the direction of $\vec{r}_{PA}$:

$\vec{F_A} = F_A \frac{\vec{r}_{PA}}{|\vec{r}_{PA}|} = 0.052 G \frac{(-0.100, -0.200)}{\sqrt{0.05}}$

$\vec{F_A} = \left( \frac{-0.052 G \times 0.100}{\sqrt{0.05}}, \frac{-0.052 G \times 0.200}{\sqrt{0.05}} \right)$

The force $\vec{F_B}$ acts in the direction of $\vec{r}_{PB}$:

$\vec{F_B} = F_B \frac{\vec{r}_{PB}}{|\vec{r}_{PB}|} = 0.052 G \frac{(0.100, -0.200)}{\sqrt{0.05}}$

$\vec{F_B} = \left( \frac{0.052 G \times 0.100}{\sqrt{0.05}}, \frac{-0.052 G \times 0.200}{\sqrt{0.05}} \right)$

5. Calculate the Net Force: The net force $\vec{F}_{net} = \vec{F_A} + \vec{F_B}$.

• X-component of net force:

$F_{net,x} = \frac{-0.052 G \times 0.100}{\sqrt{0.05}} + \frac{0.052 G \times 0.100}{\sqrt{0.05}} = 0$

(Due to symmetry, the horizontal components cancel out).

• Y-component of net force:

$F_{net,y} = \frac{-0.052 G \times 0.200}{\sqrt{0.05}} + \frac{-0.052 G \times 0.200}{\sqrt{0.05}}$

$F_{net,y} = 2 \times \left( \frac{-0.052 G \times 0.200}{\sqrt{0.05}} \right) = \frac{-0.0208 G}{\sqrt{0.05}}$

So, the net force vector is $\vec{F}_{net} = \left( 0, -\frac{0.0208 G}{\sqrt{0.05}} \right)$. The magnitude of the net force is $F_{net} = \frac{0.0208 G}{\sqrt{0.05}}$. The direction is vertically downwards.

6. Calculate the Initial Acceleration: Using Newton's Second Law, $\vec{F}_{net} = m\vec{a}$.

$\vec{a} = \frac{\vec{F}_{net}}{m} = \frac{1}{0.010 \text{ kg}} \left( 0, -\frac{0.0208 G}{\sqrt{0.05}} \right)$

$\vec{a} = \left( 0, -\frac{0.0208 G}{0.010 \sqrt{0.05}} \right)$

$\vec{a} = \left( 0, -\frac{2.08 G}{\sqrt{0.05}} \right)$

The magnitude of the acceleration is $a = \frac{2.08 G}{\sqrt{0.05}}$. To simplify the expression:

$a = \frac{2.08 G}{\sqrt{5/100}} = \frac{2.08 G}{\sqrt{5}/10} = \frac{2.08 \times 10 G}{\sqrt{5}} = \frac{20.8 G}{\sqrt{5}}$ To rationalize the denominator:

$a = \frac{20.8 \sqrt{5} G}{5} = 4.16 \sqrt{5} G \text{ m/s}^2$

The direction of the acceleration is the same as the net force, which is vertically downwards (towards the origin).

The final answer is $\boxed{4.16\sqrt{5}G \text{ m/s}^2 \text{ vertically downwards}}$.

Explanation of the solution:

1. Calculated distances from P to A and P to B, finding them equal ($r = \sqrt{0.05}$ m).
2. Calculated the magnitude of gravitational force from A (or B) on P using $F = \frac{GMm}{r^2}$, which is $0.052G$.
3. Resolved forces into components. Due to symmetry, horizontal components cancel out.
4. Summed vertical components: $F_{net,y} = -2 F \cos\theta$. The angle $\theta$ of the force vector with the vertical axis has $\cos\theta = \frac{0.2}{\sqrt{0.05}}$.
5. Calculated net force $F_{net,y} = -\frac{0.0208 G}{\sqrt{0.05}}$.
6. Calculated acceleration $a = \frac{F_{net,y}}{m}$ to get $a = \frac{20.8 G}{\sqrt{5}} = 4.16 \sqrt{5} G \text{ m/s}^2$.
7. Direction is vertically downwards.