Question 7.10: Goal Use vectors to find the net gravitational force on an o...
Goal Use vectors to find the net gravitational force on an object.
Problem (a) Three 0.300-kg billiard balls are placed on a table at the corners of a right triangle, as shown from overhead in Figure 7.19. Find the net gravitational force on the cue ball (designated as m_1) resulting from the forces exerted by the other two balls. (b) Find the components of the gravitational force of m_2 on m_3.
Strategy (a) To find the net gravitational force on the cue ball of mass m_1, we first calculate the force \overrightarrow{F}_{21} exerted by m_2 on m_1. This force is the y-component of the net force acting on m_1. Then we find the force \overrightarrow{F}_{31} exerted by m_3 on m_1, which is the x-component of the net force acting on m_1. With these two components, we can find the magnitude and direction of the net force on the cue ball. (b) In this case, we must use trigonometry to find the components of the force . \overrightarrow{F}_{23}
(a) Find the net gravitational force on the cue ball. Find the magnitude of the force \overrightarrow{F}_{21} exerted by m_2 on m_1 using the law of gravitation, Equation 7.20:
F = G\frac{m_1m_2}{r^2} [7.20]
{F}_{21} = \frac{m_2m_1}{{r_{21}}^2}
= (6.67 × 10^{-11} N·m^2/kg^2)\frac{(0.300 kg)(0.300 kg)}{(0.400 m)^2}
F_{21} = 3.75 × 10^{-11} N
Find the magnitude of the force \overrightarrow{F}_{31} exerted by m_3 on m_1, again using Newton’s law of gravity:
{F}_{31} = G\frac{m_3m_1}{{r_{31}}^2}
= (6.67 × 10^{-11} N·m^2/kg^2)\frac{(0.300 kg)(0.300 kg)}{(0.300 m)^2}
F_{31} = 6.67 × 10^{-11} N
The net force has components F_x = F_{31} and F_y = F_{21}. Compute the magnitude of this net force:
F = \sqrt{F_x^2 + F_y^2} = \sqrt{(6.67)^2 + (3.75)^2} × 10^{-11} N
= 7.65 × 10^{-11} N
Use the inverse tangent to obtain the direction of \overrightarrow{F}:
θ = tan^{-1}(\frac{F_y}{F_x}) = tan^{-1}(\frac{3.75 × 10^{-11} N}{6.67 × 10^{-11} N}) = 29.3°
(b) Find the components of the force of m_2 on m_3. First, compute the magnitude of \overrightarrow{F}_{23}:
{F}_{23} = G\frac{m_2m_1}{{r_{23}}^2}
= (6.67 × 10^{-11} kg^{-1}m^3s^{-2})\frac{(0.300 kg)(0.300 kg)}{(0.500 m)^2}
= 2.40 × 10^{-11} N
To obtain the x- and y-components of F_{23} we need cos φ and sin φ. Use the sides of the large triangle in Figure 7.19:
F_c = ma_c = m\frac{v^2}{r} [7.19]
\cos φ = \frac{adj}{hyp} = \frac{0.300 m}{0.500 m} = 0.600
\sin φ = \frac{opp}{hyp} = \frac{0.400 m}{0.500 m} = 0.800
Compute the components of \overrightarrow{F}_{23} A minus sign must be supplied for the x-component, because it’s in the negative x-direction.
F_{23x} = – F_{23} \cos φ = – (2.40 × 10^{-11} N)(0.600)
= – 1.44 × 10^{-11} N
F_{23y} = F_{23} \sin φ = (2.40 × 10^{-11} N)(0.800)
= 1.92 × 10^{-11} N
Remarks Notice how small the gravity forces are between everyday objects. Nonetheless, such forces can be measured directly with torsion balance