Question

Find the resultant of three coplanar concurrent forces; F1 acting north and of magnitude 8 grams, F2 of magnitude 12 grams and acting southwest, F3 of magnitude 5 grams and acting southeast. Find the components of the resultant in an easterly and northerly direction.

Answer 1

Answer on Question #63909, Physics / Mechanics | Relativity

Question:

Find the resultant of three coplanar concurrent forces; F1 acting north and of magnitude 8 grams, F2 of magnitude 12 grams and acting southwest, F3 of magnitude 5 grams and acting southeast. Find the components of the resultant in an easterly and northerly direction.

Solution:

In Cartesian coordinate system X-axis is equivalent to easterly direction, and Y-axis — to northerly direction. We may decompose these three forces as

$$\vec {F} _ {1} = \left(F _ {1} ^ {x}; F _ {1} ^ {y}\right) = (0; 8)$$ $$\begin{array}{l} \vec {F} _ {2} = \left(F _ {3} ^ {x}; F _ {2} ^ {y}\right) = (- 1 2 \cdot \sin \alpha ; - 1 2 \cdot \cos \alpha) = (- 1 2 \cdot \sin 4 5 {}^ {\circ}; - 1 2 \cdot \cos 4 5 {}^ {\circ}) \\ = \left(- 1 2 \cdot \frac {\sqrt {2}}{2}; - 1 2 \cdot \frac {\sqrt {2}}{2}\right) = (- 6 \sqrt {2}; - 6 \sqrt {2}) \\ \end{array}$$ $$\vec {F} _ {3} = \left(F _ {3} ^ {x}; F _ {3} ^ {y}\right) = (5 \cdot \cos \alpha ; - 5 \cdot \sin \alpha) = \left(5 \cdot \frac {\sqrt {2}}{2}; - 5 \cdot \frac {\sqrt {2}}{2}\right) = (2. 5 \sqrt {2}; - 2. 5 \sqrt {2})$$

The components of the resultant force are:

$$F _ {r e s} ^ {x} = F _ {1} ^ {x} + F _ {2} ^ {x} + F _ {3} ^ {x} = 0 - 6 \sqrt {2} + 2. 5 \sqrt {2} = - 3. 5 \sqrt {2} \cong - 4. 9 5 g r a m s$$ $$F _ {r e s} ^ {y} = F _ {1} ^ {y} + F _ {2} ^ {y} + F _ {3} ^ {y} = 8 - 6 \sqrt {2} - 2. 5 \sqrt {2} = 8 - 8. 5 \sqrt {2} \cong - 4. 0 2 g r a m s$$

Answer:

-4.95 and -4.02 grams

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