A15.0 kg sphere is at the origin and a 7.00 kg sphere is at x=20cm. at what position on the x-axis could you place a small mass such that the net gravitational force on it due to the spheres is zero?

The position of the small mass on the x- axis = 11.87 cm

Explanation:

From Newton's law of universal gravitation,

F₁ = Gm₁X/r₁² Equation 1

Where F₁ = Force exerted on the on the small mass by the first sphere, m₁ = mass of the first sphere, X = mass of the small mass, r₁ = distance between the first sphere and the small mass.

Also,

F₂ = Gm₂X/r₂² Equation 2

Where F₂ = Force exerted by the second sphere on the small mass, m₂ = mass of the second sphere, X = mass of the small mass, r₂ = distance between the second sphere and the small mass.

Note: F₁ = F₂ ( Net force on the small mass due to the sphere is zero)

Therefore,

Gm₁X/r₁² = Gm₂X/r₂²

Equating the similar terms from both side of the equation,

m₁/r₁² = m₂/r₂² Equation 3

Given: m₁ = 15.0 kg, m₂ = 7.00 kg, let r₁ = y cm, then r₂ = (20-y) cm

Substituting these values into equation 3

15/y² = 7/(20-y)²

15(20-y)² = 7y²

√{15(20-y)² = √7y²

√15×√(20-y)² = √7×√y² ( from the law of surd)

3.87(20-y) = 2.65y

77.4 - 3.87y = 2.65y

Collecting like terms and reordering the equation

2.65y+3.87y = 77.4

6.52y = 77.4

y = 77.4/6.52

y = 11.87 cm

Thus the position of the small mass on the x- axis = 11.87 cm

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever