Question

Objects with masses of 255 kg and a 555 kg are separated by 0.390 m.
(a) Find the net gravitational force exerted by these objects on a 32.0-kg object placed midway between them.
(b) At what position (other than infinitely remote ones) can the 32.0-kg object be placed so as to experience a net force of zero?

Answer 1

Answer:

(a). The net gravitational force is $4.20\times10^{-6}\ N$

(b). The position is at 0.232 m.

Explanation:

Given that,

Mass of one object M = 255 kg

Mass of another object M'= 555 kg

Separation = 0.390 m

(a). We need to calculate the net gravitational force

Using formula of force

$F_{net}=\dfrac{GM'm}{r^2}-\dfrac{GMm}{r^2}$

$F_{net}=\dfrac{Gm(M'-M)}{r^2}$

Put the value into the formula

$F_{net}=\dfrac{6.67\times10^{-11}\times32.0(555-255)}{(0.390)^2}$

$F_{net}=4.20\times10^{-6}\ N$

The net gravitational force is $4.20\times10^{-6}\ N$

(b). We need to calculate the position

Force from 555 mass = Force from 255 mass

$\dfrac{Gm\times555}{x^2}=\dfrac{Gm\times255}{(0.390-x)^2}$

$555(0.390-x)^2=255x^2$

$300x^2-0.78x+84.4155=0$

$x=0.232\ m$

The position is at 0.232 m.

Hence, (a). The net gravitational force is $4.20\times10^{-6}\ N$

(b). The position is at 0.232 m.

Answer 2

The net gravitational force exerted by the two object on 32 kg object mid way is 4.55 x 10⁻⁵ kg.

The position that the 32 kg mass will placed so as to experience a zero net force does not exist.

The given parameters;

• mass of the first object, m₁ = 255 kg
• mass of the second object, m₂ = 555 kg
• distance between the two objects, r = 0.39 m

The net gravitational force exerted by the two object on 32 kg object mid way;

$F_{net} = F_{12} + F_{23}\\\\F_{net} = \frac{Gm_1m_2}{(r/2)^2} + \frac{Gm_2m_3}{(r/2)^2}\\\\F_{net} = \frac{(6.67\times 10^{-11})(255\times 32)}{(0.39/2)^2} \ + \ \frac{(6.67\times 10^{-11})(555\times 32)}{(0.39/2)^2} \\\\F_{net} = 1.43\times 10^{-5} \ + \ 3.115 \times 10^{-5} \\\\F_{net} = 4.55 \times 10^{-5} \ N$

The position in which the 32 kg will be placed so that it experiences a zero net force is calculated as follows;

$\Sigma \ F =0\\\\\frac{Gm_1m_2}{(x)^2} + \frac{Gm_2m_3}{(0.39 -x)^2} = 0\\\\\frac{Gm_1m_2}{(x)^2} = - \frac{Gm_2m_3}{(0.39 -x)^2}\\\\-Gm_2m_3x^2 = Gm_1m_2(0.39-x)^2\\\\-m_3x^2= m_1(0.39 - x)^2\\\\-m_3x^2 = m_1(0.152 - 0.78x + x^2)\\\\-555x^2 = 255(0.152 -0.78x + x^2)\\\\-55 5 x^2 = 38.76 - 198.9 x + 255x^2\\\\-810x^2 +198.9x - 38.76 = 0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = -810, \ \ b = 198.9 \ \ c = -38.76\\\\x = \frac{-b \ \ + /-\ \ \sqrt{b^2 -4ac} }{2a}$

$x = \frac{-198.9 \ \ + /-\ \ \sqrt{(198.9)^2 -4(-810\times -38.76)} }{2(-810)} \\\\x= imaginary \ roots$

Thus, the position that the 32 kg mass will placed so as to experience a zero net force does not exist.

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