The cables attached to the screw eye are subjected


The cables attached to the screw eye are subjected to the three forces shown. Express each force in Cartesian vector form and determine the magnitude and coordinate direction angles of the resultant force.

The cables attached to the screw eye are subjected

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

We will first look at force F_1. To express it in Cartesian vector notation, we need to know the amount of force exerted, which is 350 N, and we need to know the angles between x, y, and z directions. Looking at the diagram, we see that there is a 50^0 angle (90^0-40^0=50^0) between force F_1 and the y direction, and there is a 40^0 angle between force F_1 and z direction.

 

Thus, we can write force F_1 as follows:

F_1=350\left\{\text{cos}\,50^0j\,+\text{cos}\,40^0k\right\} N

F_1=\left\{224.98j\,+\,268.12k\right\} N

(Note that we don’t have an i value because there is no angle between force F_1 and the x-axis)

 

Let us now look at force F_2. Again, looking at the diagram, we see that the force exerted is 100 N. We also see that there is a 45^0 angle between force F_2 and the x-axis, a 60^0 angle between force F_2 and the y-axis, and a 120^0 angle between force F_2 and the z-axis.

 

Thus, we can write force F_2 like this:

F_2=100\left\{\text{cos}\,45^0i\,+\,\text{cos}\,60^0j\,+\,\text{cos}\,120^0k\right\} N

F_2=\left\{70.71i\,+\,50.0j\,-\,50k\right\} N

 

Lastly, we will look at force F_3. We see that it exerts a force of 250 N. We also see by looking at the diagram, that there is a 60^0 angle between force F_3 and the x-axis, a 135^0 angle between force F_3 and the y-axis, and a 60^0 angle between force F_3 and the z-axis.

It’s important to see how the angle between force F_3 and the y-axis is 135^0. Note on the image that force F_3 is in the negative y direction. Thus, it’s actually \text{sin}\,(-45)^0 which is the same as \text{cos}\,(135)^0.

 

Thus, we can write force F_3 like so:

F_3=250\left\{\text{cos}\,60^0i\,+\,\text{cos}\,135^0j\,+\,\text{cos}\,60^0k\right\} N

F_3=\left\{125.0i\,-\,176.78j\,+\,125.0k\right\} N

 

Now that we have each force written in the Cartesian vector notation, we can easily find the resultant force in Cartesian vector notation. We do so by adding each of the forces together. In a more simple sense, we are adding all the i, j, and k vectors together individually. So we can write the resultant as:

F_R=F_1+F_2+F_3

F_R=\left\{(70.71+125.0)i\,+\,(224.98+50.0-176.78)j\,+\,(268.12-50.0+125.0)k\right\} N

(Again, we simply added each i, j, and k group together, individually.)
Now, we will simplify by adding each value inside the brackets.

F_R=\left\{195.71i\,+\,98.20j\,+\,343.12k\right\} N

 

With this, we can now find the magnitude of the resultant force. Remember that the magnitude is:

F_R=\sqrt{F^2_{R_x}\,+\,F^2_{R_y}\,+\,F^2_{R_z}}

F_R=\sqrt{(195.71)^2\,+\,(98.20)^2\,+\,(343.12)^2}

F_R=407.03 N

 

Now, all that is left is to find the coordinate direction angles.

 

\text{cos}\,\alpha=\frac{F_{R_x}}{F_R}=\frac{195.71}{407.03}

\alpha=\text{cos}^{-1}(0.4808)=61.3^0

 

\text{cos}\,\beta=\frac{F_{R_y}}{F_R}=\frac{98.20}{407.03}

\beta=\text{cos}^{-1}(0.2413)=76.0^0

 

\text{cos}\,\gamma=\frac{F_{R_z}}{F_R}=\frac{343.12}{407.03}

\gamma=\text{cos}^{-1}(0.843)=32.5^0

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-77.

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