Two spheres A and B have an equal mass 2


Two spheres A and B have an equal mass and are electrostatically charged such that the repulsive force acting between them has a magnitude of 20 mN and is directed along the line AB. Determine the angle θ, the tension in the cords AC and BC, and the mass m of each sphere.

Two spheres A and B have an equal mass

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

We will draw two free body diagrams for sphere A and B. Let us start with sphere B.

Two spheres A and B have an equal mass

Note how we got the angles. The colored orbs correspond to the alternate interior angles.

We will now write our equations of equilibrium for sphere B. First we will write our equation of equilibrium for the x-axis forces.

\rightarrow ^+\sum \text{F}_\text{x}\,=\,0

0.02\text{cos}\,(30^0)\,-\,T_{BC}\text{sin}\,(30^0)\,=\,0

(Note we converted 20 mN to 0.02 N)

Solve for T_{BC}

T_{BC}\,=\,0.0346\text{N}\,=\,34.6 mN

 

Now, we will write our equation of equilibrium for y-axis forces.

Show me the free body diagram

+\uparrow \sum \text{F}_\text{y}\,=\,0

0.02\text{sin}\,(30^0)\,+\,T_{BC}\text{cos}\,(30^0)\,-\,W\,=\,0

(Substitute the value of T_{BC} we found and solve for W)

W\,=\,0.0399 N

 
To figure out the mass, remember that W = mg

(Where W is weight, m is mass and g is the force of gravity)

m\,=\,\dfrac{W}{g}

m\,=\,\dfrac{0.0399}{9.81}

m\,=\,0.004\,\text{kg}\,=\,4 g

 

Let us now draw our free body diagram for sphere A.

Two spheres A and B have an equal mass

Again, we will write our equations of equilibrium.

\rightarrow ^+\sum \text{F}_\text{x}\,=\,0

T_{AC}\text{sin}\,(\theta)\,-\,0.02\text{cos}\,(30^0)\,=\,0 (eq.1)

 

+\uparrow \sum \text{F}_\text{y}\,=\,0

T_{AC}\text{cos}\,(\theta)\,-\,0.02\text{sin}\,(30^0)\,-W\,=\,0 (eq.2)

(remember, we already found W. W = 0.0399 N)

 

Let us now solve for T_{AC} and \theta.

Isolate for T_{AC} in eq.1.

T_{AC}=\dfrac{0.02\text{cos}\,(30^0)}{\text{sin}\,(\theta)} (eq.3)

 
Substitute this value into eq.2

\dfrac{0.02\text{cos}\,(30^0)}{\text{sin}\,(\theta)}\text{cos}\,(\theta)\,-\,0.02\text{sin}\,(30^0)\,-0.0399\,=\,0

(Simplify all possible terms)

\dfrac{0.0173\text{cos}\,(\theta)}{\text{sin}\,(\theta)}\,=\,0.0499

(Multiply both sides by \dfrac{1}{0.0173})

\dfrac{\text{cos}\,(\theta)}{\text{sin}\,(\theta)}\,=\,2.884

To solve for \theta we must remember the identity \dfrac{\text{cos}\,(\theta)}{\text{sin}\,(\theta)}\,=\,\text{cot}\,(\theta)

Therefore, we can write it as,

\text{cot}\,(\theta)\,=\,2.884

Again, we must remember the identity \text{cot}\,(\theta)\,=\,\dfrac{1}{\text{tan}\,(\theta)}

\dfrac{1}{\text{tan}\,(\theta)}\,=\,2.884

(Take the reciprocal of both sides)

\text{tan}\,(\theta)\,=\,\dfrac{1}{2.884}

\theta\,=\,\text{tan}^{-1}\left(\dfrac{1}{2.884}\right)

\theta\,=\,19.1^0

 

We can now substitute the value of θ into eq.3 to figure out T_{AC}.

T_{AC}\,=\,\dfrac{0.02\text{cos}\,(30^0)}{\text{sin}\,(19.1}

T_{AC}\,=\,0.0529\text{N}\,=\,52.9 mN

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-16.

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