Respuesta :
1) -7.14 N
2) +2.70 N
3) 7.63 N
Explanation:
1)
In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.
The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is
[tex]180^{\circ}-61^{\circ}[/tex]
so its x-component is
[tex]F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N[/tex]
F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is
[tex]180^{\circ}+52.8^{\circ}[/tex]
Therefore its x-component is
[tex]F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N[/tex]
So, the x-component of the resultant force is
[tex]F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N[/tex]
2)
In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.
The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is
[tex]180^{\circ}-61^{\circ}[/tex]
so its y-component is
[tex]F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N[/tex]
F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is
[tex]180^{\circ}+52.8^{\circ}[/tex]
Therefore its y-component is
[tex]F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N[/tex]
So, the y-component of the resultant force is
[tex]F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N[/tex]
3)
The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.
Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:
[tex]F=\sqrt{F_x^2+F_y^2}[/tex]
Where in this problem, we have:
[tex]F_x=-7.14 N[/tex] is the x-component
[tex]F_y=2.70 N[/tex] is the y-component
And substituting, we find:
[tex]F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N[/tex]
Answer:
Explanation:
F1 = 8 N, along 61° from negative x axis in Second quadrant
f2 = 5.4 N, along 52.8° from negative x axis in third quadrant
Let R be the resultant force.
X component of resultant force,
Rx = - F1 x Cos 61° - F2 Cos 52.8°
Rx = - 8 x 0.485 - 5.4 x 0.605
Rx = - 3.88 - 3.27
Rx = - 7.15 N
Y component of resultant force,
Ry = F1 x Sin 61° - F2 Sin 52.8°
Ry = 8 x 0.875 - 5.4 x 0.8
Ry = 7 - 4.32
Ry = 2.68 N
The magnitude of resultant force
[tex]R=\sqrt{R_{x}^{2}+R_{y}^{2}}[/tex]
[tex]R=\sqrt{ 7.15^{2}+2.68^{2}}[/tex]
R = 7.64 N
