COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 13.
Free-Body Diagram:
ΣM D = 0:
( 750 N )( 0.1 m − a ) − ( 750 N )( a + 0.075 m − 0.1 m ) − (125 N )( 0.05 m ) + B ( 0.2 m ) = 0
87.5 N + 0.2 B
a=
1500 N
(1)
Using the bounds on B:
B = − 250 N (i.e. 250 N downward) in (1) gives amin = 0.0250 m
B = 500 N (i.e. 500 N upward) in (1) gives amax = 0.1250 m
Therefore:
25.0 mm ≤ a ≤ 125.0 mm