Three astronauts, propelled by jet backpacks, push and guide a 131 kg asteroid toward a processing dock, exerting the forces shown in the figure, with F1 = 36 N, F2 = 58 N, F3 = 41 N, θ1 = 30°, and θ3 = 60°. What is the (a) magnitude and (b) angle (measured relative to the positive direction of the x axis in the range of (-180°, 180°]) of the asteroid's acceleration?

Three astronauts, propelled by jet backpacks, push and guide a 121 kg asteroid toward a processing dock, exerting the forces shown in the figure, with F1 = 30 N, F2 = 58 N, F3 = 45 N, è1 = 30°, and è3 = 60°. What is the (a) magnitude and (b) angle (measured relative to the positive direction of the x axis in the range of (-180°, 180°]) of the asteroid's acceleration?

To find the magnitude and angle of the asteroid's acceleration, we need to determine the net force acting on it using vector addition.

Step 1: Resolve the forces into their x and y components.
Given:
F1 = 36 N, θ1 = 30°
F2 = 58 N
F3 = 41 N, θ3 = 60°

Resolving F1 into its x and y components:
F1x = F1 * cos(θ1)
F1y = F1 * sin(θ1)

Resolving F3 into its x and y components:
F3x = F3 * cos(θ3)
F3y = F3 * sin(θ3)

Step 2: Calculate the x and y components of the net force.
The net force in the x-direction is the sum of the x-components of the individual forces:
Net Fx = F1x + F2 + F3x

The net force in the y-direction is the sum of the y-components of the individual forces:
Net Fy = F1y + F3y

Step 3: Calculate the magnitude and angle of the net force.
The magnitude of the net force can be found using the Pythagorean theorem:
Net F = √((Net Fx)^2 + (Net Fy)^2)

The angle of the net force can be found using trigonometry:
θ = arctan(Net Fy / Net Fx)

Now let's calculate the values:

Step 1:
F1x = 36 N * cos(30°) ≈ 31.18 N
F1y = 36 N * sin(30°) ≈ 18 N

F3x = 41 N * cos(60°) ≈ -20.5 N (negative direction because of the angle)
F3y = 41 N * sin(60°) ≈ 35.5 N

Step 2:
Net Fx = F1x + F2 + F3x
= 31.18 N + 58 N - 20.5 N
= 68.68 N

Net Fy = F1y + F3y
= 18 N + 35.5 N
= 53.5 N

Step 3:
Magnitude of the net force:
Net F = √((Net Fx)^2 + (Net Fy)^2)
= √((68.68 N)^2 + (53.5 N)^2)
≈ √(4712.21 N^2 + 2862.25 N^2)
≈ √7574.46 N^2
≈ 86.99 N

Angle of the net force:
θ = arctan(Net Fy / Net Fx)
= arctan(53.5 N / 68.68 N)
≈ 36.43°

(a) The magnitude of the asteroid's acceleration is approximately 86.99 N, and (b) the angle of the asteroid's acceleration is approximately 36.43° (measured relative to the positive direction of the x-axis in the range of (-180°, 180°]).

You did not give the angle for F2.