Adjustable Wrench Determine the moment about the center of the nut when the 250-N force is applied to the adjustable wrench. Use (a) the perpendicular lever arm method and (b) the moment components method. (See Figure 4.13 on page 146.) Approach We are tasked with finding the resulting moment

Question

Adjustable Wrench

Determine the moment about the center of the nut when the 250-N force is applied to the adjustable wrench. Use (a) the perpendicular lever arm method and (b) the moment components method. (See Figure 4.13 on page 146.)

Approach
We are tasked with finding the resulting moment using two methods. We again assume that we can neglect the impact of the wrench’s weight. We denote the center of the nut as point A, and the point of application of the force as point B. The moment is calculated by applying Equations (4.8)

[latex]M_{\omicron} = Fd[/latex] (4.8)

and (4.9).

[latex] M_{\omicron}=\pm F_{x}\Delta y \pm F_{y}\Delta x [/latex] (4.9)

We will use the trigonometric equations from Appendix B to determine the necessary lengths and angles. (See Figure 4.14.)

Accepted Answer

(a) We first need to determine the length of the perpendicular lever arm, and this step involves geometrical constructions. By using the given dimensions,

[latex]AB=\sqrt{(75mm)^2+(200mm)^2} \longleftarrow [z^2=x^2+y^2] [/latex]

= 213.6 mm

Although this is the distance to the location at which the force is applied, it is not the perpendicular lever arm’s length. For that, we need to calculate the length of AC, which is perpendicular to the force’s line of action. Because the force is inclined by 35° from vertical, a line perpendicular to it is oriented 35° from horizontal. Line AB lies at the angle

[latex]\alpha = an ^{-1}(\frac{75mm}{200 mm} ) \longleftarrow [ an heta =\frac{y}{x} ][/latex]
[latex]= an ^{-1}(0.375\frac{\cancel{mm}}{\cancel{mm}} )[/latex]
[latex]= an ^{-1}(0.375)=20.56^{\omicron }[/latex]

below horizontal, and so it is offset by

[latex]β=35° - 20.56° = 14.44°[/latex]

from line AC. The perpendicular lever arm distance becomes

[latex]d = (213.6 mm) \cos 14.44° \longleftarrow [x= z \cos heta ] [/latex]

= 206.8 mm

and the wrench’s moment becomes

[latex]M_{A} = (250 N)(0.2068 m) \longleftarrow [M_{A}= Fd ] [/latex]

= 51.71 N · m

which is directed clockwise.
(b) In the moment components method, the 250-N force is broken down into its horizontal and vertical components having magnitudes (250 N) sin 35° = 143.4 N and (250 N) cos 35° = 204.8 N. Those components are oriented leftward and downward in the diagram. Individually, each exerts a clockwise moment about point A. Referring to our sign convention in the figure, a counterclockwise moment is positive. By summing the moment produced by each force component, we have

[latex]M_{A} = -(143.4 N)(0.075 m) - (204.8 N)(0.2 m) \longleftarrow [M_{A}=\pm F_x\Delta y\pm F_y\Delta x ] [/latex]

= - 51.71 N · m

Because the numerical value is negative, the net moment is directed clockwise.

Discussion
In this instance, it’s probably easier to apply the moment components method because the horizontal and vertical dimensions of the wrench are given in the problem’s statement. Also, if gravity was accounted for, the clockwise moment would be larger because gravity would create an additional force acting down around the middle of the wrench.

[latex]M_{A}[/latex] = 51.71 N · m (clockwise)

Question 4.4: Adjustable Wrench Determine the moment about the center of t...

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