The box wrench in Fig. 5–15a is used to tighten the bolt at A. If the wrench does not turn when the load is applied to the handle, determine the torque or moment applied to the bolt and the force of the wrench on the bolt.

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Free-Body Diagram. The free-body diagram for the wrench is shown in Fig. 5–15 b . Since the bolt acts as a “fixed support,” it exerts force components [latex]\mathbf{A}_{ x}[/latex] and [latex]\mathbf{A}_{ y}[/latex] and a moment [latex]\mathbf{M}_{ A}[/latex] on the wrench at A .
Equations of Equilibrium.

[latex] \begin{array}{c} \stackrel{+}{ ightarrow} \Sigma F_x=0 ; \quad A_x-52\left(\frac{5}{13} ight) N +30 \cos 60^{\circ} N =0 \ A_x=5.00 N \end{array} [/latex]

[latex] \begin{array}{c} +\uparrow \Sigma F_y=0 ; \quad A_y-52\left(\frac{12}{13} ight) N -30 \sin 60^{\circ} N =0 \ A_y=74.0 N \end{array} [/latex]

[latex] \begin{array}{c} ⤿ +\Sigma M_A=0 ; M_A-\left[52\left(\frac{12}{13} ight) N ight](0.3 m )-\left(30 \sin 60^{\circ} N ight)(0.7 m )=0 \ M_A=32.6 N · m \end{array} [/latex]

Note that [latex]\mathbf{M}_{ A}[/latex] must be included in this moment summation. This couple moment is a free vector and represents the twisting resistance of the bolt on the wrench. By Newton’s third law, the wrench exerts an equal but opposite moment or torque on the bolt. Furthermore, the resultant force on the wrench is

[latex] F_A=\sqrt{(5.00)^2+(74.0)^2}=74.1 N [/latex]

NOTE: Although only three independent equilibrium equations can be written for a rigid body, it is a good practice to check the calculations using a fourth equilibrium equation. For example, the above computations may be verified in part by summing moments about point C:

[latex] ⤿ +\Sigma M_C=0 ; \quad\left[52\left(\frac{12}{13} ight) N ight](0.4 m )+32.6 N · m -74.0 N (0.7 m )=0 [/latex]

19.2 N · m + 32.6 N · m - 51.8 N · m = 0

Question 5.8: The box wrench in Fig. 5–15a is used to tighten the bolt at ...

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