Questions: Se desea sostener el cuerpo de 140 lb que se muestra en la figura. Diga qué tensión T deberá aplicarse para lograrlo y cuál debe ser el ángulo. (Sol. T=81.2 lb ; Θ=29.5°)

Solution Steps

The body is subjected to three forces:

1. The weight of the body (140 lb) acting downward.
2. A force of 80 lb acting at an angle of 30° to the horizontal.
3. The tension $T$ acting at an angle $\theta$ to the horizontal.

Resolve the 80 lb force into horizontal and vertical components:

• Horizontal component: $80 \cos(30°)$
• Vertical component: $80 \sin(30°)$

For the body to be in equilibrium, the sum of the forces in both the horizontal and vertical directions must be zero.

$$T \cos(\theta) = 80 \cos(30°)$$

$$T \sin(\theta) + 80 \sin(30°) = 140$$

First, calculate the components of the 80 lb force:

• $80 \cos(30°) = 80 \times \frac{\sqrt{3}}{2} = 40\sqrt{3} \approx 69.28$ lb
• $80 \sin(30°) = 80 \times \frac{1}{2} = 40$ lb

Using the horizontal equilibrium equation: $$T \cos(\theta) = 69.28$$

Using the vertical equilibrium equation: $$T \sin(\theta) + 40 = 140$$ $$T \sin(\theta) = 100$$

Divide the vertical equation by the horizontal equation to find $\theta$: $$\tan(\theta) = \frac{100}{69.28}$$ $$\theta = \tan^{-1}\left(\frac{100}{69.28}\right) \approx 29.5°$$

Now, solve for $T$ using $\cos(\theta)$: $$T = \frac{69.28}{\cos(29.5°)} \approx 81.2 \text{ lb}$$

Final Answer