Question

Topic Gravitational force amd firld strength.. help me please

Answer 1

The gravitational force between m₁ and m₂ has magnitude

`F_(1,2) = (Gm_1m_2)/(x^2)`

while the gravitational force between m₁ and m₃ has magnitude

`F_(1,3) = (Gm_1m_3)/((15-x)^2)`

where x is measured in m.

The mass m₁ is attracted to m₂ in one direction, and attracted to m₃ in the opposite direction such that m₁ in equilibrium. So by Newton's second law, we have

`F_(1,2) - F_(1,3) = 0`

Solve for x :

`(Gm_1m_2)/(x^2) = (Gm_1m_3)/((15-x)^2) \\\\ (m_2)/(x^2) = (m_3)/((15-x)^2) \\\\ ((15-x)^2)/(x^2) = (m_3)/(m_2) = (60\,\rm kg)/(40\,\rm kg) = \frac32 \\\\ \left(\frac{15-x}x\right)^2 = \frac32 \\\\ \left(\frac{15}x-1\right)^2 = \frac32 \\\\ \frac{15}x - 1 = \pm √(\frac32) \\\\ \frac{15}x = 1 \pm √(\frac32) \\\\ x = (15)/(1\pm√(\frac32))`

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

`x \approx 6.74\,\mathrm m`