Question

Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant. 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.
What is the x component of the resultant force?
What is the y component of the resultant force?
What is the magnitude of the resultant force?

Answer 1

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

$180^{\circ}-61^{\circ}$

so its x-component is

$F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N$

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

$180^{\circ}+52.8^{\circ}$

Therefore its x-component is

$F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N$

So, the x-component of the resultant force is

$F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N$

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

$180^{\circ}-61^{\circ}$

so its y-component is

$F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N$

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

$180^{\circ}+52.8^{\circ}$

Therefore its y-component is

$F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N$

So, the y-component of the resultant force is

$F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N$

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:

$F=\sqrt{F_x^2+F_y^2}$

Where in this problem, we have:

$F_x=-7.14 N$ is the x-component

$F_y=2.70 N$ is the y-component

And substituting, we find:

$F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N$

Answer 2

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

$R=\sqrt{R_{x}^{2}+R_{y}^{2}}$

$R=\sqrt{ 7.15^{2}+2.68^{2}}$

R = 7.64 N