If FB = 2kN and the resultant force acts along 2


If F_B = 2kN and the resultant force acts along the positive u axis, determine the magnitude of the resultant force and the angle u.

If FB = 2kN and the resultant force acts along

Solution:

Let us first draw the vector components as follows:

If FB = 2kN and the resultant force acts along

Notice that the resultant is along the u-axis as stated by the question.

Now, we will draw the vector components tail to tail as follows:

If FB = 2kN and the resultant force acts along

We can now use law of sines to figure out \phi. (Forgot the law of sines?)

\frac{\sin \phi}{3}=\frac{\sin30^0}{2}

\phi=48.69^0

From this, we can figure out \theta.

\theta=30^0+\phi

\theta = 78.59^0

 

Now we can use the law of sines again to figure out F_R.

\frac{F_R}{\sin(180^0-78.59^0)}=\frac{2}{\sin30^0}

F_R=3.92kN

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-7.

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2 thoughts on “If FB = 2kN and the resultant force acts along

    • questionsolutions Post author

      I am sure there are other ways to solve these types of questions though I haven’t utilized them. Most textbooks usually show two ways of solving these questions, either the tip to tip method or the parallelogram way. Please take a look: https://goo.gl/ssjJMP

      Many thanks and best of luck with your studies. đŸ™‚