2 An object of mass $m_1=5.00\U{kg}$ placed on a frictionless,
3 horizontal table is connected to a string that passes over a pulley
4 and then is fastened to a hanging object of mass $m_2=9.00\U{kg}$ as
5 shown in Figure~P5.28. \Part{a} Draw free-body diagrams of both
6 objects. Find \Part{b} the magnitude of that acceleration of the
7 objects and \Part{c} the tension in the string.
14 real a = u; // block side length
15 real pr = a/4; // pulley radius
16 real psw = 0.2u; // pulley support width
17 real d = 2u; // rope length
19 Surface s = Surface((-.7a, 0), (d-2.5pr, 0));
20 Block m1 = Block((0, a/2), width=a, height=a, "$m_1$");
21 Block m2 = Block(m1.center + (d,-d), width=a, height=a, "$m_2$");
23 pair ropecross = extension(m1.center, m1.center+E,
24 m2.center, m2.center+N);
25 pair pulley = ropecross + (-pr, -pr/Tan(90/2));
28 draw(pulley -- (pulley + 1.3(s.pTo-pulley)), psw+currentpen);
30 draw(m1.center -- (pulley.x, m1.center.y));
31 draw(m2.center -- (m2.center.x, pulley.y));
33 filldraw(shift(pulley)*scale(pr)*unitcircle, fillpen=white);
52 real t = (m1*m2*g)/(m1 + m2);
54 Vector T = Force((0,0), mag=t, dir=0, "$T$");
56 Vector G = Force((0,0), mag=m1*g, dir=-90, "$F_g$");
58 Vector norm = Force((0,0), mag=m1*g, dir=90, "$N$");
60 dot("$m_1$", (0,0), W);
70 real t = (m1*m2*g)/(m1 + m2);
72 Vector T = Force((0,0), mag=t, dir=90, "$T$");
74 Vector G = Force((0,0), mag=m2*g, dir=-90, "$F_g$");
76 dot("$m_2$", (0,0), E);
81 Because the rope does not stretch, both objects have the same
82 magnitude of acceleration. Using $F=ma$ on both objects, we can solve
87 m_2 g - m_1 a &= m_2 a \\
88 m_2 g &= (m_1 + m_2) a \\
89 a &= \frac{m_2 g}{m_1 + m_2}
90 = \frac{9.00\U{kg}\cdot9.80\U{m/s$^2$}}{5.00\U{kg} + 9.00\U{kg}}
91 = \ans{6.30\U{m/s$^2$}}
95 Plugging the solution for $a$ back into either of the $F=ma$
99 = \frac{m_1 m_2 g}{m_1 + m_2}
100 = \frac{5.00\U{kg}\cdot9.00\U{kg}\cdot9.80\U{m/s$^2$}}
101 {5.00\U{kg} + 9.00\U{kg}}