Objects with masses of 135 kg and a 435 kg are separated by 0.350 m. (a) find the net gravitational force exerted by these objects on a 67.0 kg object placed midway between them. magnitude n direction (b) at what position (other than infinitely remote ones) can the 67.0 kg object be placed so as to experience a net force of zero? m from the 435 kg mass

a) 3.94. 10⁻⁵ N  b) 0.23 m. from 435 Kg. mass.

Explanation:

a) In order to find the net gravitational force on the 67.0 Kg object, we need to apply the superposition principle to the forces due to gravity, exerted by m1 and m2 on the mass located between them.

These forces are explained by the Universal Gravitation Law, discovered by Isaac Newton, and can be written as follows:

F = G m₁. m₂ / (r₁₂)², which is a vector equation.

If we assume that the three masses are aligned over a horizontal line, we can write the vector equation as an algebraic one, with components only on the x axis, taking the direction towards the greater mass as the positive one, so we can write the following equation, derived from the expression for the Universal Gravitation Law.

Fnet = F1 + F2 = G. mₓ / (.35/2) m² . (405 kg - 135 Kg)

Replacing mx= 67.0 Kg, and G = 6.67 . 10⁻¹¹ N. kg⁻².m⁻², we find the gravitational net force on the 67.0 Kg, solving for Fnet:

Fnet = 6,67. 10⁻¹¹ N. Kg⁻² . m⁻². 67.0 Kg /(. 175)² m². 300 Kg = 3.94.10⁻⁵ N, directed towards the bigger mass, 405 Kg.

b) In order to find a point between both masses where the net force on the third mass be zero, we need to make equal each other the magnitude of the force exerted by m₁ and m₂, as follows:

F₁ = F₂

Now, if we call x to the distance between the bigger mass and the third mass, as we know that the distance between m₁ and m₂ is 0.35 m, the distance between m1 and the 67.0 Kg mass, will be (0.35 - x).

So, we can write the following expression:

F₁ = F₂ ⇒ G m₁ mₓ / (0.35-x)² = G m₂ mₓ / x²

Simplifying common terms, and solving the 2nd degree equation for x, we get only one point between both masses that satisfy the equation, as follows:

X = 0.23 m from the 435 Kg. mass.

dang that balls went far