Resultant force is the combined force from a system of forces. The resultant force R of two forces F1 and F2 is: R is of magnitude 73.3 N and is 19.47 degrees up from AC.
How to find resultant force of a system of forces?
Many tricks can be used, but we'll use the basic method.
Convert all forces of the system in component form.
Calculate the sum on those components.
Get the magnitude of the force by using Pythagoras theorem(which will work for force's vector space).
The given forces are:
[tex]F_1 = 38N, 50^\circ up\\F_2 = 40N, 0^\circ up[/tex]
Converting them in component forms, one horizontal and one vertical component, we get:
[tex]_iF_1 = F_1cos(\theta) = 38\times cos(50^\circ) \approx 29.11 N \\_jF_1 = F_1sin(\theta) = 38 \times sin(50^\circ) \approx 24.43N\\\\_iF_2 = F_2cos(\theta) = 40\times cos(0^\circ)= 40N \\_jF_2 = F_2sin(\theta) = 40 \times sin(0^\circ) = 0N\\\\[/tex]
Adding them component wise, we get:
[tex]_iR = \: _iF_1 + \: _iF_2= 69.11 N\\_jR = \: _jF_1 + \: _jF_2= 24.43 N\\[/tex]
Let we have R = x N, and y degrees, then:
[tex]xcos(y) = \: _iR = 69.11 N\\xsin(y) = \:_jR = 24.43 N\\\\\text{Squaring and adding}\\\\x^2(sin^2y + cos^2 y) \approx 5373.01\\\\x =\sqrt{5373.01} = 73.3 N\text{ \: Positive root since sign of force is stored in angles}\\Thus,\\\\y = cos^{-1}(\dfrac{69.11}{73.3}) \approx 19.47277471^\circ, -19.47277471^\circ\\\\y = 19.47^\circ \text{\: (As AB is above horizontal axis, thus, positive)}[/tex]
Thus, R is of magnitude 73.3 N and is 19.47 degrees up from AC.
Learn more about resultant forces here:
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