Respuesta :
Answer:
Magnitude of the resultant = 357.85 N
Direction of the resultant is 12.1° from the 220 N force and still in between the two forces 150 N and 220 N
Explanation:
Let the direction in which the 220 N force is acting be the x-axis, then the 150 N force is acting 30° from the 220 N force, i.e. the x-axis.
The 220 N force in vector form = (220cos 0°)î + (220sin 0°)j = (220î) N
The 150 N force in vector form = (150cos 30°)î + (150sin 30°)j = (129.9î + 75j) N
The resultant is the vector sum of both forces = (220î) + (129.9î + 75j) = (349.9î + 75j)
Magnitude = √(349.9² + 75²) = 357.85 N
Direction = tan⁻¹ (75/349.9) = 12.1° From the 220 N force (x-axis)
- Magnitude of the resultant of the force = 357.85 N
- Direction of the resultant is 12.1° from the 220 N force and still in between the two forces 150 N and 220 N
Before solving this question first learn some concepts about force:
- Vector quantities are quantities that possess both magnitude and direction.
- A force has both magnitude and direction, therefore: Force is a vector quantity.
- Its units are Newtons, N.
Consider the direction in which the 220 N force is acting be the x-axis, then the 150 N force is acting 30° from the 220 N force, i.e. the x-axis.
So, let's write the expression for the force in "vector form"
220 N= (220cos 0°)î + (220sin 0°)j = (220î) N
150 N = (150cos 30°)î + (150sin 30°)j = (129.9î + 75j) N
The resultant is the vector sum of both forces = (220î) + (129.9î + 75j) = (349.9î + 75j)
Magnitude = [tex]\sqrt{349.9^{2}+75^{2} } =357.28 N[/tex]
Direction = [tex]tan^{-1} =\frac{75}{349.9} =12.1^o[/tex] From the 220 N force (x-axis)
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