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A body is in equilibrium when three forces act on it. Two of the forces are F1 of magnitude 8N...

Question:

A body is in equilibrium when three forces act on it. Two of the forces are F1 of magnitude 8N acting due east and F2 which is 5N acting N 60degree E. Find the magnitude and direction of the third force.

Equilibrium of Forces:

When to or more forces are acting on a body the body remains in equilibrium if vector sum of all the forces acting on the body is zero. Here we shall find the resultant of the 8 N and 5 N forces. The magnitude of the required force will be equal to the magnitude of resultant and just opposite to the resultant.

Answer and Explanation:

The magnitude of third force 'F' is given by

{eq}F^2=F_1^2+F_2^2+2F_1F_2\cos60 \\ F^2=8^2+5^2+2 \times 8 \times 5\cos60 \\ F=11.35\ N {/eq}

The direction of this force with east is given by

{eq}tan \alpha= \dfrac{F_2 \sin 60}{F_1+F_2 cos60} \\ tan \alpha= \dfrac{5 \sin 60}{8+5 cos60} \\ \Rightarrow \alpha =22.4^\circ {/eq}

Hence the third force should make an angle of 180+22.4=200.4 degree with east

or 22.4 degree south of west


Learn more about this topic:

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Vectors: Definition, Types & Examples

from Common Entrance Test (CET): Study Guide & Syllabus

Chapter 57 / Lesson 3
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