Two forces are applied to a box on a frictionless surface. One force is 40 N [S]. The second force is 70 N [N30ºW]. What is the overall net force?

a) 40.6 N [S59.5ºE]
b) 110 N
c) 40.6 N [N59.5ºW]
d) 40.6 N [S30.5ºW]

To find the overall net force, we need to add the two forces together. However, since the forces are not in the same direction, we will need to break down the second force into its horizontal and vertical components.

The first force is given as 40 N [S]. This means it is a 40 N force pointing towards the south.

The second force is given as 70 N [N30ºW]. This force has two components: a vertical component pointing north and a horizontal component pointing west.

To determine the magnitude of the vertical component, we can use trigonometry. The angle between the force vector and the north direction is 30º. The sine of this angle gives us the proportion of the force that is pointing north.

sin(30º) = vertical component / 70 N
vertical component = sin(30º) * 70 N ≈ 35 N

Therefore, the vertical component of the force is 35 N [N].

To determine the magnitude of the horizontal component, we can again use trigonometry. The angle between the force vector and the west direction is 30º. The cosine of this angle gives us the proportion of the force that is pointing west.

cos(30º) = horizontal component / 70 N
horizontal component = cos(30º) * 70 N ≈ 60.62 N

Therefore, the horizontal component of the force is 60.62 N [W].

Now that we have the vertical and horizontal components of the second force, we can add them to the first force to find the overall net force.

In the vertical direction, we have:
40 N [S] + 35 N [N] = 5 N [S]

In the horizontal direction, we have:
60.62 N [W]

Combining these components, the overall net force is approximately 5 N [S] + 60.62 N [W].

To find the resultant force vector, we can use the Pythagorean theorem to calculate the magnitude and trigonometry to calculate the angle.

Magnitude:
√((5 N)^2 + (60.62 N)^2) ≈ 61.20 N

Angle:
tan^(-1)((5 N) / (60.62 N)) ≈ 4.67º

Therefore, the overall net force is approximately 61.20 N [S4.67ºW].

Comparing this with the answer choices, the correct option is not provided. Therefore, none of the given options accurately represents the overall net force.

40 S = <0,-40>

70 N30ºW = <-35,60.6>

add them up, then convert back to compass directions.