August 23, 2016 in Mechanics: Statics tagged engineering Mechanics: Statics / rectangular components
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Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and also Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

You are watching: Determine the coordinate direction angle α of the resultant force.


Solution:

We will begin off by illustration a vector diagram to aid us visualize the steps far better like so:

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Note that us only emphasis on force F_1 since the question just asks us to recognize the coordinate direction angle of pressure F_1. Thus, we only attracted the vector materials of pressure F1. No need to draw unnecessary components because that will just make that more an overwhelming for us to visualize the steps.

Let us currently look in ~ the materials of force F_1.

We can write every component prefer so:

(F_1)_x=600(\frac45)\,\textcos\,30^0 N

(F_1)_y=600(\frac45)\,\textsin\,30^0 N

(F_1)_x=600(\frac35) N

We can now write pressure F_1 in Cartesian vector type like so:

F_1=600\left\\frac45\,\textcos\,30^0(+i)\,+\,\frac45\,\textsin\,30^0(-j)\,+\,\frac35(+k)\right\

F_1=600\left<0.6928i-0.4j+0.6k\right> N

(the indications for i,j, and k deserve to be checked out looking at the diagram. We have the right to see the the j ingredient is in the an adverse direction, thus, we have the right to write it as -j)

Now, we require to uncover the unit vector. To perform so, remember the we merely divide the Cartesian vector type by the value of F_1 which in this case is 600 N.

(u_F)_1=\frac600(0.6928i-0.4j+0.6k)\,\textN600\,\textN

(u_F)_1=0.6928i-0.4j+0.6k

The last action is to find the coordinate direction angles.

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\alpha=\textcos^-1((u_F)_1)_x

\alpha=\textcos^-1(0.6928) =46.1^0

\beta=\textcos^-1((u_F)_1)_y

\beta=\textcos^-1(-0.4) =114^0

\gamma=\textcos^-1((u_F)_1)_z

\gamma=\textcos^-1(0.6) =53.1^0

This question deserve to be uncovered in engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-75.