Magnitude of the resultant force FR = F1 + F2 8


Determine the magnitude of the resultant force F_{R} = F_{1} + F_{2} and its direction, measured clockwise from the positive u axis.

Magnitude of the resultant force FR = F1 + F2

Solution:

Let us draw the vector components. When doing this question, it is highly recommended to draw a very large diagram, otherwise, the v-axis can cause confusion as it might appear to be the resultant force.

Magnitude of the resultant force FR = F1 + F2

Now, we will use the law of cosines to figure out F_R. (Forgot the Law of Cosines?)

(F_{R})^2=300^2+500^2-2(300)(500)\cos95^0

F_R=\sqrt{300^2+500^2-2(300)(500)\cos95^0}

F_R=605.1N

 

To figure out \theta we will use the law of sines. (Forgot the Law of Sines?)

\dfrac{605.1}{\sin95^0}=\dfrac{500}{\sin \theta}

\theta = 55.40^0

 

As the question asks us for the direction measured from the positive u-axis, we need to add 30^0 to our \theta value. Remember that \theta only represents the angle between force F_1 and F_R.

\phi=55.40^0+30^0

\phi=85.4^0

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-4.

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