Question 3.SP.1: A 100-lb vertical force is applied to the end of a lever, wh......
A 100-lb vertical force is applied to the end of a lever, which is attached to a shaft at O. Determine (a) the moment of the 100-lb force about O; (b) the horizontal force applied at A that creates the same moment about O; (c) the smallest force applied at A that creates the same moment about O; (d) how far from the shaft a 240-lb vertical force must act to create the same moment about O; (e) whether any one of the forces obtained in parts b, c, or d is equivalent to the original force.
STRATEGY: The calculations asked for all involve variations on the basic defining equation of a moment, M_O = Fd.
- A 100-lb vertical force is applied to the end of a lever.
- The lever is attached to a shaft at O.
- The force creates a moment about O.
- The lever is not specified in terms of length or dimensions.
- A horizontal force is applied at point A.
- The horizontal force creates the same moment about O.
- The magnitude of the horizontal force at A is unknown.
- The smallest force applied at A that creates the same moment about O is unknown.
- A 240-lb vertical force is applied at an unknown distance from the shaft.
- The 240-lb vertical force creates the same moment about O.
- The distance from the shaft where the 240-lb vertical force must act to create the same moment about O is unknown.
- It is unclear whether any of the forces obtained in parts b, c, or d are equivalent to the original force.
Final Answer
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 \text { in. }) \cos 60^{\circ}=12 \text { in. }
The magnitude of the moment about O of the 100-lb force is
M_O=F d=(100~ \mathrm{lb})(12~ \mathrm{in} .)=1200~ \mathrm{lb} \cdot \mathrm{in} .
Since the force tends to rotate the lever clockwise about O, represent the moment by a vector \mathbf{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 \mathrm{lb} \cdot \mathrm{in} \circlearrowright
b. Horizontal Force. In this case, you have (Fig. 2)
d=(24 \text { in. }) \sin 60^{\circ}=20.8 \text { in. }
Since the moment about O must be 1200 lb·in., you obtain
\begin{aligned}M_O & =F d \\1200~ \mathrm{lb} \cdot \text {in. } & =F(20.8~ \mathrm{in} .) \\F & =57.7~ \mathrm{lb} \quad \mathbf{F}=57.7~ \mathrm{lb} \rightarrow\end{aligned}
c. Smallest Force. Since 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
\begin{aligned}M_O & =F d \\1200 ~\mathrm{lb} \cdot \text {in. } & =F(24 \text { in. }) \\F & =50~ \mathrm{lb} \quad \mathbf{F}=50 ~\mathrm{lb} \measuredangle 30^{\circ}\end{aligned}
d. 240-lb Vertical Force. In this case (Fig. 4), M_O = Fd yields
1200~ \mathrm{lb}\cdot \text { in. }=(240~ \mathrm{lb}) d \quad d=5~ \mathrm{in} .
but
O B \cos 60^{\circ}=d
so O B=10~ \mathrm{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.
