2 Fig. P4.24 shows loads hanging from the ceiling of an elecator that is
3 moving at a constant velocity. Find the tension in each of the three
4 strands of cord supporting each load.
5 \end{problem*} % problem 4.24
8 First, we need to understand the effect of the elevator.
9 It is moving at a constant {\it velocity} so we know that
10 the acceleration $\vect{a}$ of all the elements must be $0$.
11 So the elevator's constant motion has no effect on the tensions.
14 Let $m = 5.00\U{kg}$ be the mass of the ball,
15 $\theta_1 = 40.0^o$ be the angle between $\vect{T}_1$ and the horizontal,
16 and $theta_2 = 50.0^o$ be the angle between $\vect{T}_2$ and the horizontal.
17 Following identical reasoning to Problem 22 \Part{a}, we know
19 $T_3 = mg = 5.00\U{kg} \cdot 9.8\U{m/s}^2 = \ans{49\U{N}}$.
21 Now looking at the knot where the three cords come together.
22 There are three forces acting on the knot: $T_1$, $T_2$, and $T_3$.
23 Letting the upwards direction be $+\vect{x}$
24 and the rightwards direction be $+\vect{y}$ we can break our tensions
27 T_{1x} &= -T_1 \cos \theta_1 \\
28 T_{1y} &= T_1 \sin \theta_1 \\
29 T_{2x} &= T_2 \cos \theta_2 \\
30 T_{2y} &= T_2 \sin \theta_2 \\
34 Now summing the forces on the knot we have
36 \sum F_x &= T_{3x} + T_{2x} + T_{1x}
37 = 0 + T_2 \cos \theta_2 - T_1 \cos \theta_1 \\
39 T_2 &= T_1 \frac{\cos \theta_1}{\cos \theta_2} \\
40 \sum F_y &= T_{3y} + T_{2y} + T_{1y}
41 = -mg + T_2 \sin \theta_2 + T_1 \sin \theta_1 \\
42 T_2 &= \frac{mg - T_1 \sin \theta_1}{\sin \theta_2}
43 = T_1 \frac{\cos \theta_1}{\cos \theta_2} \\
44 \frac{mg}{\cos \theta_1} - T_1 \frac{sin \theta_1}{\cos \theta_1}
45 &= T_1 \frac{\sin \theta_2}{\cos \theta_2} \\
46 T_1 (\tan \theta_1 + \tan \theta_2) &= \frac{mg}{\cos \theta_1} \\
47 T_1 &= \frac{mg}{\cos \theta_1 (\tan \theta_1 + \tan\theta_2)}
48 = \frac{49\U{N}}{\cos 40^o (\tan 40^o + \tan 50^o)}
50 T_2 &= T_1 \frac{\cos \theta_1}{\cos \theta_2}
51 = \frac{mg}{\cos \theta_2 (\tan \theta_1 + \tan\theta_2)}
52 = \frac{49\U{N}}{\cos 50^o (\tan 40^o + \tan 50^o)}
57 The only changes from \Part{a} are
58 $m = 10\U{kg}$, $\theta_1 = 60.0^o$, and $\theta_2 = 0^o$.
59 Plugging the new values into our symbolic equation from \Part{a}:
61 T_3 &= mg = 10.0\U{kg} \cdot 9.8\U{m/s}^2 = \ans{98\U{N}} \\
62 T_1 &= \frac{mg}{\cos \theta_1 (\tan \theta_1 + \tan\theta_2)}
63 = \frac{98\U{N}}{\cos 60^o (\tan 60^o + \tan 0^o)}
65 T_2 &= T_1 \frac{\cos \theta_1}{\cos \theta_2}
66 = \frac{mg}{\cos \theta_2 (\tan \theta_1 + \tan\theta_2)}
67 = \frac{98\U{N}}{\cos 0^o (\tan 60^o + \tan 0^o)}