The plate is subjected to the two forces at A 6


The plate is subjected to the two forces at A and B as shown.If \theta =60^0,determine the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal.

The plate is subjected to the two forces at A

Solution:

Let us draw the vector components as follows:

The plate is subjected to the two forces at A

Now, let us draw the vectors tail to tail as follows:

The plate is subjected to the two forces at A

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

(F_R)^2=8^2+6^2-2(8)(6)\cos100^0

F-R=\sqrt{8^2+6^2-2(8)(6)\cos100^0}

F_R=10.80 kN

Now, we can calculate \theta by using law of sines. (Forgot the Law of Sines?)

\frac{\sin \theta}{6}=\frac{\sin 100^0}{10.80}

\sin \theta = 0.5470

\theta = 33.16^0

Now, to calculate \phi we simply subtract 30^0 from \theta

\phi=33.16^0-30^0

\phi=3.16^0

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-15.

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