2 Two objects are connected by a light string that passes over a
3 frictionless pulley as shown in Figure~P5.30. Assume the incline is
4 frictionless and take $m_1=2.00\U{kg}$, $m_2=6.00\U{kg}$, and
5 $\theta=55.0\dg$. \Part{a} Draw free-body diagrams of both objects.
6 Find \Part{b} the magnitude of the acceleration of the
7 objects, \Part{c} the tension in the string, and \Part{d} the speed of
8 each object $2.00\U{s}$ after it is released from rest.
15 real theta = 55.0; // ramp angle in degrees
16 real w = 2u; // width of ramp base
21 pair top = (0, w*Tan(theta));
23 real a = u; // diameter of m_1, side length of m_2
24 real dx = 0.1u; // narrow sliver separating m_1 from wedge wall
25 real pr = a/4; // pulley radius
26 real psw = 0.2u; // pulley support width
28 Surface s = Surface((botL - (0.2w + a + dx, 0)), botR + (0.2w, 0));
30 Angle A = Angle(botL, botR, top, "$\theta$");
32 Mass m1 = Mass(top - (a/2 + dx, 1.5a), radius=a/2, "$m_1$");
33 pair m2bot = (top+botR)/2;
34 pair m2perp = rotate(-90)*dir(top-botR);
36 m2bot + a/2*m2perp, width=a, height=a, direction=-theta, "$m_2$");
38 pair ropecross = extension(m1.center(), m1.center()+N,
39 m2.center, m2.center+(top-botR));
40 pair pulley = ropecross + (pr, -pr/Tan(theta/2));
45 draw(pulley -- (pulley + 1.3(top-pulley)), psw+currentpen);
47 filldraw(botL -- botR -- top -- cycle, fillpen=rgb(0.8,0.8,0.3));
50 draw(m1.center() -- (m1.center().x, pulley.y));
51 draw(m2.center -- (pulley + pr*m2perp));
53 filldraw(shift(pulley)*scale(pr)*unitcircle, fillpen=white);
72 real t = g*m1*m2/(m1+m2)*(1+Sin(theta));
74 Vector T = Force((0,0), mag=t, dir=90,
75 L=Label("$T$", position=EndPoint, align=E));
77 Vector G = Force((0,0), mag=m1*g, dir=-90,
78 L=Label("$m_1 g$", position=EndPoint, align=E));
80 dot("$m_1$", (0,0), W);
91 real t = g*m1*m2/(m1+m2)*(1+Sin(theta));
93 Vector T = Force((0,0), mag=t, dir=180-theta,
94 L=Label("$T$", position=EndPoint, align=NE));
96 Vector N = Force((0,0), mag=m2*g*Cos(theta), dir=90-theta,
97 L=Label("$N$", position=EndPoint, align=E));
99 Vector G = Force((0,0), mag=m2*g, dir=-90,
100 L=Label("$m_2 g$", position=EndPoint, align=E));
102 dot("$m_2$", (0,0), SE);
107 Because the rope does not stretch, both objects have the same
108 magnitude of acceleration. Using $F=ma$ on both objects, we can solve
109 for $a$. $m_2$ is the heavier object, so we'll pick the positive
110 direction to be dropping $m_2$ and raising $m_1$.
112 T - m_1 g &= m_1 a \\
113 m_2 g \sin(\theta) - T &= m_2 a \\
115 m_2 g \sin(\theta) - m_1 (g+a) &= m_2 a \\
116 g (m_2\sin(\theta) - m_1) &= a (m_1 + m_2) \\
117 a &= g \frac{m_2\sin(\theta)-m_1}{m_1 + m_2} \\
118 &= 9.80\U{m/s$^2$}\cdot\frac{6.00\U{kg}\cdot\sin(55.0\dg)-2.00\U{kg}}
119 {2.00\U{kg} + 6.00\U{kg}} \\
120 &= \ans{3.57\U{m/s$^2$}}
124 Plugging the solution for $a$ back into either of the $F=ma$
128 = m_1 g \p({1 +\frac{m_2\sin(\theta)-m_1}{m_1 + m_2}})
129 = m_1 g \frac{m_1 + m_2 + m_2\sin(\theta)-m_1}{m_1 + m_2} \\
130 &= m_1 g \frac{m_2(1+\sin(\theta))}{m_1 + m_2}
131 = g \frac{m_1 m_2}{m_1 + m_2}(1+\sin(\theta)) \\
132 &= 9.80\U{m/s$^2$}\cdot\frac{2.00\U{kg} \cdot 6.00\U{kg}}
133 {2.00\U{kg} + 6.00\U{kg}}
134 \cdot(1+\sin(55.0\dg))
139 Because the string does not stretch, the speed of both objects are the
140 same. Because the acceleration is constant,
142 v = a\cdot t + v_0 = a\cdot t = 3.57\U{m/s$^2$} \cdot 2.00\U{s}