Question

the beam shown in (figure 1) has a weight of 1200 lb. part a determine the shortest cable abc that can be used to lift it if the maximum force the cable can sustain is 1500 lb. express your answer in feet to three significant figures. l_abc =

Explanation:

Step1: Analyze force equilibrium

Let the tension in the cable be $T$. When the beam is on the verge of being lifted, the maximum - force the cable can sustain $T = 1500$ lb. The weight of the beam $W=1200$ lb. Consider the equilibrium of forces. Let the vertical component of the tension in the cable support the weight of the beam.

Step2: Use trigonometry

Let the length of the cable be $L_{ABC}$. Assume the height of the beam is $h = 10$ ft. Let the vertical component of the tension $T_y$ supports the weight of the beam. So $T_y=W$. If the angle between the cable and the vertical is $\theta$, then $T_y = T\cos\theta$. So $\cos\theta=\frac{W}{T}=\frac{1200}{1500}=0.8$. Then $\sin\theta=\sqrt{1 - \cos^{2}\theta}=\sqrt{1 - 0.8^{2}} = 0.6$.

Step3: Calculate cable length

The vertical height from the bottom - most point of the beam to the top - most point of the beam is $h = 10$ ft. If we consider the right - triangle formed by the cable and the vertical, and let the length of the cable be $L_{ABC}$, then using the relationship $\cos\theta=\frac{h}{L_{ABC}}$. So $L_{ABC}=\frac{h}{\cos\theta}$. Substituting $h = 10$ ft and $\cos\theta = 0.8$, we get $L_{ABC}=\frac{10}{0.8}=12.5$ ft.

Answer:

$12.5$ ft