Problem 3.102 The mass of block A is 42 kg, and the
mass of block B is 50 kg. The surfaces are smooth. If
the blocks are in equilibrium, what is the force F?
B
F
45°
A
20°
Solution: Isolate the top block. Solve the equilibrium equations.
The weight is. The angle between the normal force N1 and the positive x axis is. The normal force is. The force N2 is. The equilibrium
conditions are
"
from which
"
"
Solve:
y
B
N2
F D N1 C N2 C W D 0
N1
Fx D (0.7071jN1 j % jN2 j'i D 0
Fy D (0.7071jN1 j % 490.5'j D 0.
N1 D 693.7 N,
W
α
x
y
N1
jN2 j D 490.5 N
Isolate the bottom block. The weight is
W D 0i % jWjj D 0i % (42'(9.81'j D 0i % 412.02j (N).
The angle between the normal force N1 and the positive x axis is
(270° % 45° ' D 225° .
F
β
N3
α
A
x
W
The normal force:
N1 D jN1 j(i cos 225° C j sin 225° ' D jN1 j(%0.7071i % 0.7071j'.
The angle between the normal force N3 and the positive x-axis is
(90° % 20° ' D 70° .
The normal force is
N1 D jN3 j(i cos 70° C j sin 70° ' D jN3 j(0.3420i C 0.9397j'.
The force is . . . F D jFji C 0j. The equilibrium conditions are
"
F D W C N1 C N3 C F D 0,
from which:
"
"
Fx D (%0.7071jN1 j C 0.3420jN3 j C jFj'i D 0
Fy D (%0.7071jN1 j C 0.9397jN3 j % 412'j D 0
For jN1 j D 693.7 N from above:
jFj D 162 N
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