The pail and its contents have a mass of 60 kg. If the cable is 15 m long, determine the distance *y* of the pulley for equilibrium. Neglect the size of the pulley at A.

#### Solution:

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Before we start the question, note the following:

- The pulley is frictionless
- The size of the pulley is negligible
- The tension in the cable is the same throughout (because the pulley is frictionless)

With these factors in mind, we can draw a free body diagram like so:

Note how the tension, T, is the same on both sides. Again, this is because we are assuming the pulley to be frictionless.

We can now write an equation of equilibrium for the x-axis forces.

T\text{cos}\theta\,-\,T\text{cos}\phi\,=\,0

(We can cancel out the T by factoring it out and dividing both sides of the equation by T)

Therefore, we are left with:

\text{cos}\theta\,=\,\text{cos}\phi

Which means:

We can now shift our focus into figuring out *y* as we established both angles to be the same. Note that to solve this problem, you must remember the concepts of similar triangles and ratios.

Let us draw a diagram showing the lengths. How we got the values will be explained below.

Looking at the initial diagram from the question, we can see that the total horizontal length is 10 m. Therefore, if we choose one side from the bucket to the be *x*, then the other side is *10 – x*.

As for the vertical values, it is given to us that the right side height is *y*, and there is a difference of 2 m between pin C and B (pay careful heed to the initial diagram to see what we mean). Therefore, the height on the left side is y - 2.

Pythagorean theorem can be used to figure out the hypotenuse of each triangle that is formed which are shown on top of the hypotenuse.

We can now write separate equations to figure out y.

First, we know that the total length of the rope is 15 m (given to us in the question). Therefore, both lengths of the hypotenuse added together must equal 15 m. We can write this as:

We can also use the concept of similar triangles (because both angles are the same) to write a ratio expression as follows:

Let us simply eq.1 by multiplying both sides by \frac{\sqrt{x^2\,+\,y^2}}{\sqrt{x^2\,+\,y^2}}.

Looking at eq.2, we see that \dfrac{10-x}{x}\,=\,\dfrac{\sqrt{(10\,-\,x)^2\,+\,(y\,-\,2)^2}}{\sqrt{x^2\,+\,y^2}} which means:

\sqrt{x^2\,+\,y^2}\left(\dfrac{10-x}{x}\right)\,+\,\sqrt{x^2\,+\,y^2}\,=\,15 (eq.3)

(factor out \sqrt{x^2\,+\,y^2})

\sqrt{x^2\,+\,y^2}\left(\dfrac{10\,-\,x}{x}\,+\,1\right)\,=\,15

(simplify)

\sqrt{x^2\,+\,y^2}\left(\dfrac{10\,-\,x\,+\,x}{x}\right)\,=\,15

(divide both sides by \sqrt{x^2\,+\,y^2})

\dfrac{10}{x}\,=\,\dfrac{15}{\sqrt{x^2\,+\,y^2}}

(solve for x)

x\,=\,0.89y (eq.4)

Now, we can shift our focus to eq.2. Remember that:

(Isolate for x)

x\,=\,\dfrac{5y}{y\,-\,1}(Substitute the value we found from eq.4)

0.89y\,=\,\dfrac{5y}{y\,-\,1}(multiply both sides by *y-1*)

0.89y(y\,-\,1)\,=\,5y

(expand)

0.89y^2\,-\,0.89y-5y=0

0.89y^2\,-\,5.89y\,=\,0

(divide both sides by *y*)

(solve for *y*)

y\,=\,6.6 m

#### Final Answer: