Objects with masses of 160 kg and a 460 kg are separated by 0.440 m. (a) Find the net gravitational force exerted by these objects on a 51.0 kg object placed midway between them. magnitude N direction -Select- B (b) At what position (other than infinitely remote ones) can the Sko kg object be placed so as to experience a net force of zero? m from the 460 kg mass
Answers
\[ F_{\text{net}} = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
Where:
- \(G\) is the gravitational constant (\(6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2\))
- \(m_1\) and \(m_2\) are the masses of the two objects (160 kg and 460 kg)
- \(r\) is the separation between the objects (0.440 m)
Plugging in the values:
\[ F_{\text{net}} = \frac{(6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2) \cdot (160 \, \text{kg}) \cdot (460 \, \text{kg})}{(0.440 \, \text{m})^2} \]
\[ F_{\text{net}} \approx 5.1 \, \text{N} \]
The direction of the net force is along the line connecting the two masses.
(b) To find the position where a 51.0 kg object can be placed so as to experience a net force of zero, we can set up an equation using the gravitational forces from each mass. Let's assume the 51.0 kg object is placed at a distance \(x\) from the 460 kg mass. The net force can be zero if the gravitational force from the 160 kg mass is equal and opposite to the gravitational force from the 460 kg mass:
\[ \frac{G \cdot m_{\text{obj}} \cdot m_{160}}{x^2} = \frac{G \cdot m_{\text{obj}} \cdot m_{460}}{(0.440 - x)^2} \]
Where:
- \(m_{\text{obj}}\) is the mass of the object (51.0 kg)
- \(m_{160}\) is the mass of the 160 kg object
- \(m_{460}\) is the mass of the 460 kg object
- \(x\) is the distance from the 460 kg mass
Solving for \(x\):
\[ \frac{m_{160}}{x^2} = \frac{m_{460}}{(0.440 - x)^2} \]
\[ m_{160} \cdot (0.440 - x)^2 = m_{460} \cdot x^2 \]
Solving this quadratic equation will give you the position \(x\) where the net force on the 51.0 kg object is zero.
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