Determine the magnitude and direction θ of the equilibrium force FAB exerted along link AB by the tractive apparatus shown. The suspended mass is 10 kg. Neglect the size of the pulley at A.

#### Solution:

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Let us first figure out the tension in the rope. As the pulleys are frictionless,** the tension along the rope is always constant at any point**. We know that the suspended mass weighs 10 kg, therefore, the tension in the cable is:

T\,=\,98.1 N

We can now draw a free body diagram around pulley A.

Let us now write our equations of equilibrium.

F_{AB}\text{cos}\theta\,-\,98.1\text{cos}\,(75^0)\,-\,98.1\text{cos}\,(45^0)\,=\,0

(simplify)

F_{AB}\text{cos}\theta\,=\,94.7 (eq.1)

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

98.1\text{sin}\,(75^0)\,-\,F_{AB}\text{sin}\theta\,-\,98.1\text{sin}\,(45^0)\,=\,0

(simplify)

F_{AB}\text{sin}\theta\,=\,25.4 (eq.2)

Let us now solve for F_{AB} and \theta by isolating for F_{AB} in eq.1.

(Substitute this value into eq.2)

\dfrac{94.7}{\text{cos}\theta}\text{sin}\theta\,=\,25.4(Divide both sides by 94.7)

\dfrac{\text{sin}\theta}{\text{cos}\theta}\,=\,0.268(Remember the identity which says \text{tan}\theta\,=\,\dfrac{\text{sin}\theta}{\text{cos}\theta})

\text{tan}\theta\,=\,0.268(solve for \theta)

\theta\,=\,\text{tan}^{-1}(0.268)

\theta\,=\,15^0

Substitute the \theta value we found back into eq.3 to figure out F_{AB}.

F_{AB}\,=\,\dfrac{94.7}{\text{cos}\,(15^0)}

F_{AB}\,=\,98 N

#### Final Answers:

F_{AB}\,=\,98 N

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