Problem 15.26 The mass of each box is 12 kg, and the
coefficient of kinetic friction between the boxes and the
surface is µk = 0.05. The system is released from rest.
Determine the magnitude of the velocity of the weights
when they have moved 1 m.
30°
45°
Solution: The free body diagrams are as shown. The normal
mg
forces are
NA = mg cos 45◦ = (12)(9.81) cos 45◦ = 83.2 N,
◦
mg
A
fA
◦
NB = mg cos 30 = (12)(9.81) cos 30 = 101.9 N,
The principle of work and energy for each box is
45°
B
fB 30°
NA
NB
(mg sin 45◦ − 0.05NA − T )(1) = 12 mv 2
(T − mg sin 30◦ − 0.05NB )(1) = 12 mv 2
T
Solving these equations, we obtain
T = 71.5 N
v = 1.12 m/s.
T
mg
mg
0.05 NA
NA
NB
0.05 NB