Question 3.SP.1: A 100-lb vertical force is applied to the end of a lever, wh......

STRATEGY: The calculations asked for all involve variations on the basic defining equation of a moment, M_{O} = Fd.
MODELING and ANALYSIS:
a. Moment about O. The perpendicular distance from O to the line of action of the 100-lb force (Fig. 1) is

d = ( 24 in. ) cos 60° = 12 in.

The magnitude of the moment about O of the 100-lb force is

M_{O} = Fd = ( 100 lb )( 12 in. ) = 1200 lb·in.

Because the force tends to rotate the lever clockwise about O, represent the moment by a vector M_{O} perpendicular to the plane of the figure and pointing into the paper. You can express this fact with the notation

M_{O} = 1200 lb·in. ↻ ◂

b. Horizontal Force. In this case, you have (Fig. 2)

d = (24 in.) sin 60° = 20.8 in.

Because the moment about O must be 1200 lb·in., you obtain

M_{O}= Fd
1200 lb·in. = F( 20.8 in. )
F = 57.7 lb                                       F = 57.7 lb → ◂

c. Smallest Force. Because M_{O} = Fd, the smallest value of F occurs when d is maximum. Choose the force perpendicular to OA and note that d = 24 in. (Fig. 3); thus,

M_{O}= Fd
1200 lb·in. = F( 24 in. )
F = 50 lb                            F = 50 lb ⦪30° ◂

d. 240-lb Vertical Force. In this case (Fig. 4), M_{O} = Fd yields

1200 lb·in. = ( 240 lb )d               d = 5 in.

but

OB cos 60° = d

so

OB = 10 in.  ◂

e. None of the forces considered in parts b, c, or d is equivalent to the original 100-lb force. Although they have the same moment about O, they have different x and y components. In other words, although each force tends to rotate the shaft in the same direction, each causes the lever to pull on the shaft in a different way.

REFLECT and THINK: Various combinations of force and lever arm can produce equivalent moments, but the system of force and moment produces a different overall effect in each case.