Answer to Question #306980 in Mechanics | Relativity for Macy

Question

3. A traffic light weighing 1 X10² N hangs from a vertical cable tied to two other cables that are fastened to a support as shown in the figure. The upper cable makes angles of 37° and 53° with the horizontal. Find the tension in the three cables.

4. The distance between two telephone poles is 50 m. When a 1 kg bird lands on the telephone wire midway between the poles, the wire sags 0.2 m. Draw a free-body

diagram of the bird. How much tension does the bird produce in the wire? Ignore the weight of the wire.

Answer 1

Explanations & Calculations

3.

• Vertical cable has the tension (say $\small T_1$) equal to the weight of the light so $\small T_1 = 100\,N$.
• At the joint there is the equilibrium of forces so that horizontal equilibrium yeilds

$\qquad\qquad \begin{aligned} \small T_2 \cos37&=\small T_3\cos 53\\ \small T_2&=\small 0.754T_3\cdots\cdots(2) \end{aligned}$

• Similarly the vertical equilibrium yeilds

$\qquad\qquad \begin{aligned} \small T_2\sin37+T_3\sin53&=\small T_1=100\\ \small T_2+1.327T_3&=\small 166.16\cdots\cdots(2) \end{aligned}$

• from (2) in (3),

$\qquad\qquad \begin{aligned} \small T_3&=\small 79.8\,N \end{aligned}$

• This result in (2),

$\qquad\qquad \begin{aligned} \small T_2&=\small 60.2\,N \end{aligned}$

4.

$\qquad\qquad \begin{aligned} \small R&=\small mg\\ \small 2T\cos\theta&=\small R=mg\\ \small T&=\small \frac{mg}{2\cos\theta}\\ &=\small \frac{1\,kg\times9.8\,ms^{-2}}{2\times0.00799}\\ &=\small 613.3\,N \end{aligned}$