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: Pearchild, 2013.

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


Solution:

We will begin off by drawing a vector diagram to assist us visualize the actions much better prefer so:

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Note that we just emphasis on force F_1 because the question only asks us to identify the coordinate direction angles of pressure F_1. Hence, we just attracted the vector components of force F1. No must attract unnecessary components considering that that will certainly only make it even more hard for us to visualize the steps.

Let us now look at the components of force F_1.

We have the right to create each component favor so:

(F_1)_x=600(frac45), extcos,30^0 N

(F_1)_y=600(frac45), extsin,30^0 N

(F_1)_x=600(frac35) N

We have the right to currently compose force F_1 in Cartesian vector create favor so:

F_1=600left\frac45, extcos,30^0(+i),+,frac45, extsin,30^0(-j),+,frac35(+k) ight

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

(the indications for i,j, and k have the right to be seen looking at the diagram. We can watch that the j component is in the negative direction, therefore, we deserve to compose it as -j)

Now, we must uncover the unit vector. To carry out so, remember that we ssuggest divide the Cartesian vector form by the value of F_1 which in this case is 600 N.

(u_F)_1=frac600(0.6928i-0.4j+0.6k), extN600, extN

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

The last action is to discover the coordinate direction angles.

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alpha= extcos^-1((u_F)_1)_x

alpha= extcos^-1(0.6928) =46.1^0

eta= extcos^-1((u_F)_1)_y

eta= extcos^-1(-0.4) =114^0

gamma= extcos^-1((u_F)_1)_z

gamma= extcos^-1(0.6) =53.1^0

This question have the right to be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-75.