# Exerted along link AB by the tractive apparatus shown 1

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.

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

#### 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:

$T\,=\,(10)(9.81)$

$T\,=\,98.1$ N

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

Let us now write our equations of equilibrium.

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

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

(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.

$F_{AB}\,=\,\dfrac{94.7}{\text{cos}\theta}$ (eq.3)

(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 $F_{AB}$.

$F_{AB}\,=\,\dfrac{94.7}{\text{cos}\theta}$

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

$F_{AB}\,=\,98$ N

$\theta\,=\,15^0$

$F_{AB}\,=\,98$ N

## One thought on “Exerted along link AB by the tractive apparatus shown”

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