Determining an Unknown Vector Magnitude The cables A and B in Fig. 2.19 exert forces FA and FB on the hook. The magnitude of FA is 100 lb. The tension in cable B has been adjusted so that the total force FA + FB is perpendicular to the wall to which the hook is attached. (a) What is the magnitude

Question

Determining an Unknown Vector Magnitude

The cables A and B in Fig. 2.19 exert forces [latex]F_{A}[/latex] and [latex]F_{B}[/latex] on the hook. The magnitude of [latex]F_{A}[/latex] is 100 lb. The tension in cable B has been adjusted so that the total force [latex]F_{A} + F_{B}[/latex] is perpendicular to the wall to which the hook is attached.

(a) What is the magnitude of [latex]F_{B}[/latex]?

(b) What is the magnitude of the total force exerted on the hook by the two cables?

Strategy
The vector sum of the two forces is perpendicular to the wall, so the sum of the components parallel to the wall equals zero. From this condition we can obtain an equation for the magnitude of [latex]F_{B}[/latex] .

Accepted Answer

(a) In terms of the coordinate system shown in Fig. a, the components of [latex]F_{A}[/latex] and [latex]F_{B}[/latex] are

[latex]F_{A} = \left|F_{A} \right| \sin 40°i + \left|F_{A} \right|\cos 40°j[/latex],

[latex] F_{B} = \left|F_{B} \right| \sin 20°i - \left|F_{B} \right|\cos 20°j[/latex].

The total force is

[latex]F_{A} + F_{B} = \left(\left|F_{A}\right| \sin 40° + \left|F_{B}\right| \sin 20° \right)i + \left(\left|F_{A}\right| \cos 40° - \left|F_{B}\right| \cos 20°\right)j. [/latex]

By setting the component of the total force parallel to the wall (the y component) equal to zero,

[latex]\left|F_{A}\right| \cos 40° - \left|F_{B}\right| \cos 20° = 0 [/latex],

we obtain an equation for the magnitude of [latex] F_{B}[/latex] :

[latex]\left|F_{B}\right| = \frac{\left|F_{A}\right| \cos 40°}{ \cos 20°} = \frac{(100) \cos 40°}{ \cos 20°} = 81.5 lb [/latex] .

(b) Since we now know the magnitude of [latex] F_{B}[/latex], we can determine the total force acting on the hook:

[latex]F_{A} + F_{B} = \left(\left|F_{A}\right| \sin 40° + \left|F_{B}\right| \sin 20°\right)i = \left\lgroup (100) \sin 40°+ (81.5)\sin 20°\right\rgroup i = 92.2i (lb) [/latex].

The magnitude of the total force is 92.2 lb.

Discussion
We can obtain the solution to (a) in a less formal way. If the component of the total force parallel to the wall is zero, we see in Fig. (a) that the magnitude of the vertical component of [latex]F_{A} [/latex] must equal the magnitude of the vertical component of [latex]F_{B} [/latex]:

[latex]\left|F_{A}\right| \cos 40° = \left|F_{B}\right| \cos 20° [/latex].

Therefore the magnitude of [latex]F_{B} [/latex] is

[latex]\left|F_{B}\right| = \frac{\left|F_{A}\right| \cos 40°}{ \cos 20°} = \frac{(100) \cos 40°}{ \cos 20°} = 81.5 lb [/latex].

Question 2.6: Determining an Unknown Vector Magnitude The cables A and B i...

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