Cable Tie-Down The eyebolt is fastened to a thick base plate, and it supports three steel cables with tensions 150 lb, 350 lb, and 800 lb.Determine the resultant force that acts on the eyebolt by using the vector algebra approach. The unit vectors i and j are oriented with the x-y coordinates as

Question

Cable Tie-Down

The eyebolt is fastened to a thick base plate, and it supports three steel cables with tensions 150 lb, 350 lb, and 800 lb. Determine the resultant force that acts on the eyebolt by using the vector algebra approach. The unit vectors i and j are oriented with the x-y coordinates as shown. (See Figure 4.6.)

Approach
We are tasked to find the resultant force on the eyebolt. By using Equations (4.1) [latex]F=F_{x}i+F_{y}j [/latex] (4.1)

and Equations (4.2)

[latex]F_{y}=F \sin heta [/latex] (4.2)

we will break each force down into its horizontal and vertical components and write them in vector form. Then we will add the respective components of the three forces to fi nd the resultant’s components. Given those, the magnitude and angle of action of R follow from Equation (4.7).

[latex] heta = an ^{-1}(\frac{R_y}{R_x} ) [/latex] (4.7)

Accepted Answer

The components of the 800-lb force are

[latex] F_{x,1}= (800 lb)\cos 45^\omicron \leftarrow[F_{x}=F \cos heta] [/latex]

= 565.7 lb

[latex] F_{y,1}= (800 lb)\sin 45^\omicron \leftarrow[F_{y}=F \sin heta] [/latex]

= 565.7 lb

and [latex]F_{1}[/latex] is written in vector form as

[latex] F_1 = 565.7i + 565.7j lb \leftarrow [F=F_{x} i+F_{y}j] [/latex]

By using the same procedure for the other two forces,

[latex] F_{2} = -(350 \sin 20°) i + (350 \cos 20°) j lb[/latex]

[latex]= - 119.7i + 328.9 j lb[/latex]

[latex] F_{3}= - 150 i lb [/latex]

To calculate the components of the resultant, the horizontal and vertical
components of the three forces are summed separately:

[latex]R_x = 565.7 - 119.7 - 150 lb \longleftarrow [/latex] [latex][R_x=\sum\limits_{i=1}^{N}{F_{x,i}} ][/latex]

= 296.0 lb

[latex]R_y = 565.7 + 328.9 \longleftarrow [/latex] [latex][R_y=\sum\limits_{i=1}^{N}{F_{y,i}} ][/latex]

= 894.6 lb

The magnitude of the resultant force is

[latex]R=\sqrt{(296.0 lb)^2 + (894.6 lb)^2} \longleftarrow [R=\sqrt{R^{2}_{x}+R^{2}_{y} } ][/latex]

=942.3 lb

and it acts at the angle

[latex] heta = an ^{-1}\left(\frac{894.6\cancel{lb}}{296.0\cancel{lb}} ight) \longleftarrow [ heta =tan^{-1}(\frac{R_y}{R_x} )][/latex]

[latex]= an ^{-1}(3.022\frac{lb}{lb} )[/latex]

[latex]= an ^{-1}(3.022)[/latex]

[latex]= 71.69^\omicron [/latex]

which is measured counterclockwise from the x-axis. (See Figure 4.7.)

Discussion
The resultant force is larger than any one force, but less than the sum of the three forces because a portion of [latex]F_{x,3} [/latex] cancels [latex]F_{x,1} [/latex]. The resultant force acts upward and rightward on the bolt which is expected. The three forces acting together place the bolt in tension by pulling up on it, as well as bending it through the sideways loading [latex]R_{x}[/latex]. As a double-check on dimensional consistency when calculating the resultant’s angle, we note that the pound units in the argument of the inverse tangent function cancel. Since[latex] R_{x}[/latex] and[latex] R_{y}[/latex] are positive, the tip of the resultant vector lies in the first quadrant of the x-y plane.

[latex]R = 942.3 lb[/latex]
[latex]θ= 71.69°[/latex] counterclockwise from the x-axis

Question 4.1: Cable Tie-Down The eyebolt is fastened to a thick base plate...

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