Question

Two forces, F⃗ 1 and F⃗ 2, act at a point. F⃗ 1 has a magnitude of 9.40 N and is directed at an angle of 58.0 ∘ above the negative x axis in the second quadrant. F⃗ 2 has a magnitude of 6.20 N and is directed at an angle of 53.1 ∘ below the negative x axis in the third quadrant. A)What is the x component of the resultant force? B)What is the y component of the resultant force? C)What is the magnitude of the resultant force?

Answer 1

Answer 2

To solve this problem, we can use vector addition. We can break each force into its x and y components using trigonometry, and then add the components to find the resultant force.

A) The x component of F⃗ 1 is given by:

F⃗ 1,x = F⃗ 1 cos(58.0°)

F⃗ 1,x = (9.40 N) cos(58.0°)

F⃗ 1,x = 4.53 N

The x component of F⃗ 2 is given by:

F⃗ 2,x = F⃗ 2 cos(-53.1°)

F⃗ 2,x = (6.20 N) cos(-53.1°)

F⃗ 2,x = 3.17 N

The x component of the resultant force is:

Fₓ = F⃗ 1,x + F⃗ 2,x

Fₓ = 4.53 N + 3.17 N

Fₓ = 7.70 N

B) The y component of F⃗ 1 is given by:

F⃗ 1,y = F⃗ 1 sin(58.0°)

F⃗ 1,y = (9.40 N) sin(58.0°)

F⃗ 1,y = 7.92 N

The y component of F⃗ 2 is given by:

F⃗ 2,y = F⃗ 2 sin(-53.1°)

F⃗ 2,y = (6.20 N) sin(-53.1°)

F⃗ 2,y = -4.94 N

The y component of the resultant force is:

Fᵧ = F⃗ 1,y + F⃗ 2,y

Fᵧ = 7.92 N - 4.94 N

Fᵧ = 2.98 N

C) The magnitude of the resultant force is given by:

F = sqrt(Fₓ^2 + Fᵧ^2)

F = sqrt((7.70 N)^2 + (2.98 N)^2)

F = 8.28 N

Therefore, the x component of the resultant force is 7.70 N, the y component of the resultant force is 2.98 N, and the magnitude of the resultant force is 8.28 N.