Magnitude of the resultant force is to be 500 N 4


If the magnitude of the resultant force is to be 500 N, directed along the positive y axis, determine the magnitude of force F and its direction θ.

Magnitude of the resultant force is to be 500 N

 

Solution:

Let us first draw the vector components out. Note that the resultant force is along the positive y-axis as stated in the question.

Magnitude of the resultant force is to be 500 N

Now, let us draw the vector components tail to tail.

2

Using this diagram, we can figure out F. To do so, we will use the law of cosines.

F^2=500^2+700^2-2(500)(700)\cos105^0

F=\sqrt{500^2+700^2-2(500)(700)\cos105^0}

F=959.78N

Now, we will use the law of sines to figure out θ.

\dfrac{\sin(90^0+ \theta)}{700}=\dfrac{\sin105^0}{959.78}\theta=45.2^0

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-3.

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