2 A bag of cement whose weight is $F_g$ hangs in equilibrium from three
3 wires shown in Figure~P5.24. Two of the wires make angles
4 $\theta_1=60.0\dg$ and $\theta_2=40.0\dg$ with the horizontal.
5 Assuming the system is in equilibrium, show that the tension in the
8 T_1 = \frac{F_g \cos\theta_2}{\sin(\theta_1+\theta_2)}
18 real d = 3u; // contact point seperation
19 real a = u; // block side length
21 pair c1 = (0,0); // left contact point
22 pair c2 = c1 + (d,0); // right contact point
23 pair c3 = extension(c1, c1+dir(-theta1),
24 c2, c2+dir(180+theta2));
26 Surface s = Surface(c2+(0.2d,0), c1-(0.2d,0));
27 Angle a1 = Angle(c2, c1, c3, "$\theta_1$");
28 Angle a2 = Angle(c1, c2, c3, "$\theta_2$");
29 Block cement = Block(c3-(0, a), width=a, height=a, "$F_g$");
34 draw(c3 -- cement.center);
38 label("$T_1$", (c1+c3)/2, SW);
39 label("$T_2$", (c2+c3)/2, SE);
40 label("$T_3$", c3-(0,a/4), E);
46 Balancing forces on the cement bag, $T_3=F_g$.
48 Balancing forces on the wire joint is a bit mor complicated and
49 deserves a free body diagram.
59 real T1 = T3 * Cos(theta2)/Sin(theta1+theta2);
60 real T2 = T3 * Cos(theta1)/Sin(theta1+theta2);
62 Angle a1 = Angle(dir(180-theta1), (0,0), (-1,0), "$\theta_1$");
63 Angle a2 = Angle(dir(theta2), (0,0), (1,0), "$\theta_2$");
64 Angle A1 = Angle(dir(180-theta1), (0,0), dir(180+theta2), radius=10mm,
65 "$\theta_1+\theta_2$");
66 Angle A3 = Angle((0,-1), (0,0), dir(-90+theta2), "$\theta_2$");
68 Vector F1 = Force((0,0), mag=T1, dir=180-theta1, "$\vect{F}_1$");
69 Vector F2 = Force((0,0), mag=T2, dir=theta2,
70 L=Label("$\vect{F}_2$", position=EndPoint, align=N));
71 Vector F3 = Force((0,0), mag=T3, dir=-90, "$\vect{F}_3$");
77 draw((-7mm,0)--(7mm,0));
78 draw((12mm*dir(-90+theta2))--(0,0)--(12mm*dir(180+theta2)));
84 draw_ijhat((1.5u,-.5u), theta2);
87 Where we've chosen a coordinate system such that $\vect{F}_2$ has no
90 Balancing forces in the $y$ direction,
92 0 &= T_1 \sin(\theta_1+\theta_2) - T_3\cos(\theta_2) \\
93 T_1 &= T_3\frac{\cos(\theta_2)}{\sin(\theta_1+\theta_2)}
94 = F_g\frac{\cos(\theta_2)}{\sin(\theta_1+\theta_2)} \;,
96 which is what we set out to show.