A crate is pulled to the right with a force of 80 N, to the left with a force of 125.8 N, upward with a force of 615.4 N, and downward with a force of 248 N.

Net external force in the x direction:
-45.8
Net external force in the y direction: 367.4 N

how do i find the magnitude of the net
external force on the object?

also...

What is the direction of the net external force on the crate (as an angle between −180◦ and 180◦, measured from the positive x axis with
counterclockwise positive)?

sqrt(Fx^2+Fy^2)

tan A = Fy/Fx

i got the magnitude i still don't get tan A = Fy/Fx

can you explain thoroughly?

To find the magnitude of the net external force on the object, you can use the Pythagorean theorem:

Magnitude of the net external force = √((Net external force in the x direction)² + (Net external force in the y direction)²)

Magnitude of the net external force = √((-45.8 N)² + (367.4 N)²)

Magnitude of the net external force = √(2101.64 N² + 134904.76 N²)

Magnitude of the net external force = √137006.4 N²

Magnitude of the net external force ≈ 370.36 N

To find the direction of the net external force on the crate, you can use inverse tangent (arctan) function. The formula is:

Direction (angle) = arctan(Net external force in the y direction / Net external force in the x direction)

Direction (angle) = arctan(367.4 N / -45.8 N)

Direction (angle) ≈ arctan(-8.014)

Direction (angle) ≈ -82.79°

Therefore, the direction of the net external force on the crate is approximately -82.79° measured counterclockwise from the positive x-axis.

To find the magnitude of the net external force on the object, you need to use the components of the forces acting on the crate. The net force can be calculated using the Pythagorean theorem, given that the net force is the square root of the sum of the squares of the individual force components:

Net force = √(Net external force in the x direction)^2 + (Net external force in the y direction)^2

In this case, the given values are:

Net external force in the x direction = -45.8 N
Net external force in the y direction = 367.4 N

Now let's substitute these values into the equation:

Net force = √((-45.8)^2 + (367.4)^2)

Calculating this, the magnitude of the net external force on the object is approximately 371.8 N.

To determine the direction of the net external force on the crate, you can use the inverse tangent (arctan) function. The direction can be found using the following equation:

Direction = arctan(Net external force in the y direction / Net external force in the x direction)

Plugging in the given values:

Direction ≈ arctan(367.4 N / -45.8 N)

Calculating this expression will give you the direction of the net external force measured from the positive x-axis counterclockwise.