Q72P The system in Fig. 12-77 is in e... [FREE SOLUTION]

91影视

Chapter 12: Q72P (page 352)

The system in Fig. 12-77 is in equilibrium. The angles are $胃_{1} = 60 掳$ and $胃_{2} = 20 掳$ , and the ball has mass $M = 2 .0 鈥塳驳$ . What is the tension in (a) string ab and (b) string bc?

Short Answer

Expert verified

The tension in the string $ab$ isrole="math" localid="1661342549047" $15 N$ .
The tension in the string $bc$ is $29 N$ .

Step by step solution

Understanding the given information

The inclination angle of $T_{1}$ is $胃_{1} = 60 掳$ .

The inclination angle of $T_{2}$ is $胃_{2} = 20 掳$ .

The mass of the ball, $M = 2.0 鈥塳驳$ .

Concept and formula used in the given question

We draw the free body diagram. The system is at equilibrium. For such a system, the vector sum of the forces acting on it is zero. We can apply this concept along the x and y axes separately and find out $T_{1}$ and $T_{2}$ ,

Formulae:

$危 {\overset{鈫�}{F}}_{net} = 0$

(a) Calculation for the tension in string ab

According to the free body diagram, you can apply static equilibrium conditions along the x-axis as

$\begin{matrix} 危 {\overset{鈫�}{F}}_{x, net} = 0 \\ T_{1} \cos 胃_{1} 鈭� T_{2} \cos 胃_{2} = 0 \\ T_{1} \cos 胃_{1} = T_{2} \cos 胃_{2} \\ T_{1} = T_{2} \frac{\cos 胃_{2}}{\cos 胃_{1}} 鈥夆赌夆赌夆赌夆赌夆赌� (1) \end{matrix}$

You can apply static equilibrium conditions along the y-axis as

$\begin{matrix} 危 {\overset{鈫�}{F}}_{y, net} = 0 \\ T_{1} \sin 胃_{1} 鈭� T_{2} \sin 胃_{2} 鈭� Mg = 0 \\ (T_{2} \frac{\cos 胃_{2}}{\cos 胃_{1}}) \sin 胃_{1} 鈭� T_{2} \sin 胃_{2} 鈭� Mg = 0 \\ T_{2} (\cos 胃_{2} \tan 胃_{1} 鈭� \sin 胃_{2}) 鈭� Mg = 0 \\ T_{2} (\cos 胃_{2} \tan 胃_{1} 鈭� \sin 胃_{2}) = Mg \\ (T_{2} \frac{\cos 胃_{2}}{\cos 胃_{1}}) \sin 胃_{1} 鈭� T_{2} \sin 胃_{2} 鈭� Mg = 0 \\ T_{2} = \frac{Mg}{(\cos 胃_{2} \tan 胃_{1} 鈭� \sin 胃_{2})} \\ T_{2} = \frac{2.0 鈥塳驳脳 9.8 {鈥尘/蝉}^{2}}{(\cos 20 掳 \tan 60 掳鈭� \sin 20 掳)} \\ T_{2} = 15 鈥塏 \end{matrix}$