A tug of war has started over a popular toy. One child pulls with a force of 2.0*10^1 N[W], a second child pulls with a force of 15N[N] and a third child pulls with a force of 4.0*10^1 N [30 S of E] calculate the net force on the toy

Add the north and east components separately.

To calculate the net force on the toy, we need to find the vector sum of the forces applied by each child. The net force is the combination of both the magnitude and direction of these forces.

Let's start by representing each force as a vector using the given information:

Child 1: 2.0 * 10^1 N[W]
Child 2: 15 N[N]
Child 3: 4.0 * 10^1 N [30 S of E]

First, we need to break down Child 3's force into its horizontal and vertical components. Since the force is 30° south of east, we can calculate the individual components using trigonometry.

Horizontal Component:
Cos(30°) = adjacent/hypotenuse
Adjacent = Cos(30°) * 4.0 * 10^1 N
Adjacent = (0.866) * 4.0 * 10^1 N
Adjacent ≈ 34.6 N[E]

Vertical Component:
Sin(30°) = opposite/hypotenuse
Opposite = Sin(30°) * 4.0 * 10^1 N
Opposite = (0.5) * 4.0 * 10^1 N
Opposite = 20 N[S]

Now we have:

Child 1: 2.0 * 10^1 N[W]
Child 2: 15 N[N]
Child 3: 34.6 N[E] + 20 N[S]

To calculate the net force, we need to add the horizontal and vertical forces separately.

Horizontal Net Force = Child 1 + Child 3 horizontal
Horizontal Net Force = 2.0 * 10^1 N[W] + 34.6 N[E]

Now we need to choose a reference direction for positive force. Let's consider the right direction (east) to be positive.

Horizontal Net Force = 2.0 * 10^1 N[W] - 34.6 N[E]
Horizontal Net Force ≈ -14.6 N[E]

Vertical Net Force = Child 2 + Child 3 vertical
Vertical Net Force = 15 N[N] + 20 N[S]

Now we need to choose a reference direction for positive force. Let's consider the upward direction (north) to be positive.

Vertical Net Force = 15 N[N] - 20 N[S]
Vertical Net Force ≈ -5 N[S]

Finally, we can calculate the magnitude and direction of the net force using these horizontal and vertical components.

Net Force = √((Horizontal Net Force)^2 + (Vertical Net Force)^2)
Net Force = √((-14.6 N[E])^2 + (-5 N[S])^2)

Net Force ≈ √(213.16 N^2 + 25 N^2)
Net Force ≈ √238.16 N^2
Net Force ≈ 15.43 N

To determine the direction of the net force, we can use trigonometry:

θ = arctan(Vertical Net Force / Horizontal Net Force)
θ = arctan((-5 N[S]) / (-14.6 N[E]))

θ ≈ 19.7° south of east

Therefore, the net force acting on the toy is approximately 15.43 N in a direction 19.7° south of east.