2 Two blocks connected by a rope of negligable mass are being dragged by
3 a horizontal force (Fig.~P5.47). Suppose $F=68.0\U{N}$,
4 $m_1=12.0\U{kg}$, $m_2=18.0\U{kg}$, and the coefficient of kinetic
5 friction between each block and the surface is $0.100$. \Part{a} Draw
6 a free-body diagram for each block. Determine \Part{b} the
7 acceleration of the system and \Part{c} the tension $T$ in the rope.
13 real h = u; // height of blocks
14 real a = u; // width of blocks
15 real d = 2a; // distance between block centers
16 real f = a; // magnitude of force
18 Surface s = Surface((-0.7a,0), (d+0.7a+f,0));
20 Block b1 = Block((0, h/2), width=a, height=h, "$m_1$");
21 Block b2 = Block((d, h/2), width=a, height=h, "$m_2$");
22 draw(b1.center -- b2.center); // rope
23 label("$T$", (b1.center+b2.center)/2, N);
26 Vector F = Force(b2.center+(a/2,0), mag=f, dir=0, "$\vect{F}$");
39 real vscale = 0.1; // rescale vertical forces
45 real t = f*m1/(m1+m2);
47 Vector G = Force((0,0), mag=m1*g*vscale, dir=-90,
48 L=Label("$m_1g$", position=EndPoint, align=S));
49 Vector N = Force((0,0), mag=m1*g*vscale, dir=90,
50 L=Label("$\vect{N}_1$", position=EndPoint, align=NE));
51 Vector T = Force((0,0), mag=t, dir=0,
52 L=Label("$T$", position=EndPoint, align=NE));
53 Vector F = Force((0,0), mag=mu*m1*g, dir=180,
54 L=Label("$\vect{F}_{f1}$", position=EndPoint, align=W));
60 dot("$m_1$", (0,0), NE);
67 real vscale = 0.1; // rescale vertical forces
73 real t = f*m1/(m1+m2);
76 Vector G = Force((0,0), mag=m2*g*vscale, dir=-90,
77 L=Label("$m_2g$", position=EndPoint, align=S));
78 Vector N = Force((0,0), mag=m2*g*vscale, dir=90,
79 L=Label("$\vect{N}_2$", position=EndPoint, align=NE));
80 Vector E = Force((0,0), mag=f, dir=0,
81 L=Label("$\vect{F}$", position=EndPoint, align=NE));
82 Vector T = Force((0,dy), mag=t, dir=180,
83 L=Label("$T$", position=EndPoint, align=NW));
84 Vector F = Force((0,-dy), mag=mu*m2*g, dir=180,
85 L=Label("$\vect{F}_{f2}$", position=EndPoint, align=SW));
92 dot("$m_2$", (0,0), NE);
97 Because the string does not stretch, the blocks will have the same
98 acceleration, and can be treated as a single block.
100 F - \mu (m_1 + m_2) g &= (m_1 + m_2)a \\
101 a + \mu g &= \frac{F}{m_1 + m_2} \\
102 a &= \frac{F}{m_1 + m_2} - \mu g \\
103 &= \frac{68.0\U{N}}{12.0\U{kg}+18.0\U{kg}}
104 - 0.100\cdot 9.80\U{m/s$^2$}
105 = \ans{1.29\U{m/s$^2$}}
109 We can use the horizontal force on $m_1$ to calculate the tension
111 T - \mu m_1 g &= m_1 a \\
113 = m_1 \frac{F}{m_1 + m_2}
114 = F \frac{m_1}{m_1 + m_2} \\
115 &= 68.0\U{N} \frac{12.0\U{kg}}{12.0\U{kg} + 18.0\U{kg}}