Respuesta :
Answer:
a) F = 72.75 N and b) T = 239º
Explanation:
In that exercise they ask us for the resultant force, the forces are vectors, for which their components are due. Let's start by writing the four forces in the xy chain system
A = 33.3N i ^
B = 30.3 N j ^
C = -70.7 N i ^
D = -92.7N j ^
Let's make the sum of each component
Fₓ = A + C
Fₓ = 33.3 -70.7
Fₓ = -37.4 N
[tex]F_{y}[/tex] = B + D
[tex]F_{y}[/tex] = 30.3 -92.7
[tex]F_{y}[/tex] = -62.4 N
Let's use the Pythagorean theorem and trigonometry to find the magnitude and angle
F² = Fx² + Fy²
F =√ [(-37.4² + (-62.4)² ] = √ 5292.52
F = 72.75 N
tan θ₁ = [tex]F_{y}[/tex] / Fₓ
tan θ₁ = -62.4 / (- 37.42) = 1.6676
θ₁ = 59º
Since the magnitudes of the two components of the force are negative, the angle is in the third quadrant, so to measure the x-axis, we must add 180º to the angle
T = 180 + 59
T = 239º
The magnitude of the net force on the object is 72.7N. While its direction is 59 degrees clockwise.
RESOLUTION OF FORCES
Force is a vector quantity. It has both magnitude and direction. The resolution of forces can be resolve to x - axis and y - axis.
Given that four forces act on an object, A = 33.3 N east, B = 30.3 N north, C = 70.7 N west, and D = 92.7 N south.
(a) The magnitude of the net force on the object can be calculate by resolving the forces into X and Y component.
X - component
X = 33.3 - 70.7
X = - 37.4 N
Y - Component
Y = 30.3 - 92.7
Y = - 62.4 N
Net Force = [tex]\sqrt{X^{2} + Y^{2} }[/tex]
Net Force = [tex]\sqrt{37.4^{2} + 62.4^{2} }[/tex]
Net Force = [tex]\sqrt{5292.52}[/tex]
Net force = 72.7 N
(b) The direction of the force can be calculated by the below formula
Tan Ф = Y/X
Tan Ф = 62.4 / 37.4
Tan Ф = 1.67
Ф = [tex]Tan^{-1}[/tex] ( 1.67 )
Ф = 59 degrees.
Therefore, the magnitude of the net force on the object is 72.7N. While its direction is 59 degrees clockwise.
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