2) Two boxes are connected by a cord running over a pulley as top of the table; the coefficient of kinetic friction between bo (a) Draw the free-body diagrams for the two boxes, identify (b) Calculate the tension in the cord. FIGURE 4-5

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On the second image is the formulas. Use the formulas to solve the problem on the first image.
2) Two boxes are connected by a cord running over a pulley as shown in Fig. 4-5. Box I of mass 8.0 kg rest on the
top of the table; the coefficient of kinetic friction between box I and the table is 0.10. Box II has a mass of 15.0 kg.
(a) Draw the free-body diagrams for the two boxes, identifying all of the forces acting on each of the masses.
(b) Calculate the tension in the cord.
||
FIGURE 4-5
Transcribed Image Text:2) Two boxes are connected by a cord running over a pulley as shown in Fig. 4-5. Box I of mass 8.0 kg rest on the top of the table; the coefficient of kinetic friction between box I and the table is 0.10. Box II has a mass of 15.0 kg. (a) Draw the free-body diagrams for the two boxes, identifying all of the forces acting on each of the masses. (b) Calculate the tension in the cord. || FIGURE 4-5
UA I U! mass mj initially rest
T0n Detween box I and the table is Mk. Box II has a mass of
Diuw the free-body diagrams for the two boxes, identifying all of the forces acting on each of the masses.
(b) Calculate the tension in the cord.
Mr FN
b)
F- Mx FN =m, a o
FN-mig = 0 O >
Fr-m,g=m.(-a) O
FIGURE 4–5
FN = m,g, plug into ☺
combine
Fr - Mx Mig =mia
elimination FT
method
- mag =m;(-a)
m2 (-a)
-MEmig +m29%3DM, + Mza
Mymig+mag = a
mitmz
plng into
Fr = Mag-
- Mkm.g+m.g
M2 .
mitmz
the normal force due to the table
Transcribed Image Text:UA I U! mass mj initially rest T0n Detween box I and the table is Mk. Box II has a mass of Diuw the free-body diagrams for the two boxes, identifying all of the forces acting on each of the masses. (b) Calculate the tension in the cord. Mr FN b) F- Mx FN =m, a o FN-mig = 0 O > Fr-m,g=m.(-a) O FIGURE 4–5 FN = m,g, plug into ☺ combine Fr - Mx Mig =mia elimination FT method - mag =m;(-a) m2 (-a) -MEmig +m29%3DM, + Mza Mymig+mag = a mitmz plng into Fr = Mag- - Mkm.g+m.g M2 . mitmz the normal force due to the table
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