Question

A 30-lb body is suspended by two cords, one making 53o with the vertical and the other making 30o with the horizontal. What are the tensions in the cord?

Answer 1

Answer on Question #63907, Physics / Mechanics | Relativity

Question:

A 30-lb body is suspended by two cords, one making $53{}^{\circ}$ with the vertical and the other making $30{}^{\circ}$ with the horizontal. What are the tensions in the cord?

Solution:

The body is in equilibrium, so we may write that $\vec{T}_1 + \vec{T}_2 + \vec{W} = \vec{0}$ .

Now we decompose this vector equation into two scalar: $T_1^x = T_2^x$ and $T_1^y + T_2^y = W$ .

$T_{1}^{x} = T_{1}\cdot \sin \alpha ,T_{2}^{x} = T_{2}\cdot \cos \beta$ and then $T_{1}\cdot \sin \alpha = T_{2}\cdot \cos \beta$ ...(1)

$T_{1}^{y} = T_{1}\cdot \cos \alpha ,T_{2}^{y} = T_{2}\cdot \sin \beta$ and $T_{1}\cdot \cos \alpha +T_{2}\cdot \sin \beta = W$ ...(2)

From equation (1) $T_{2} = T_{1}\cdot \frac{\sin\alpha}{\cos\beta}$ , and we substitute it into equation (2):

$T_{1}\cdot \cos \alpha +T_{1}\cdot \frac{\sin\alpha}{\cos\beta}\cdot \sin \beta = W$ $\triangleright T_{1}\cdot (\cos \alpha +\sin \alpha \cdot \tan \beta) = W$

Finally $T_{1} = \frac{W}{\cos\alpha + \sin\alpha\cdot\tan\beta} = \frac{mg}{\cos\alpha + \sin\alpha\cdot\tan\beta}$

and $T_{2} = \frac{mg}{\cos\alpha + \sin\alpha\cdot\tan\beta}\cdot \frac{\sin\alpha}{\cos\beta} = \frac{mg\sin\alpha}{\cos\alpha\cos\beta + \sin\alpha\sin\beta} = \frac{mg\sin\alpha}{\cos(\alpha - \beta)}$

$m = 30lb = 30\cdot 0.45kg = 13.5kg$

$g = 9.81m / s^2$ $\alpha = 53{}^{\circ}$ $\beta = 30{}^{\circ}$

$T_{1} = \frac{13.5\cdot 9.81}{\cos 53{}^{\circ} + \sin 53{}^{\circ}\cdot\tan 30{}^{\circ}} = 124.6N$

$T_{2} = \frac{13.5\cdot 9.81\cdot\sin 53{}^{\circ}}{\cos(53{}^{\circ} - 30{}^{\circ})} = 114.9N$

Answer: $T_{1} = 124.6N$ $T_{2} = 114.9N$

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