Two children playing on the beach are pulling on an inner tube. One exerts a force of 45 N [N]. The other exerts a force of 60 N [SW]. What is the net force acting on the tube?
Break the second force into a S and a W vector. The S subtracts from the original N vector. With that resultant, add to the W vector.
To find the net force acting on the tube, we need to combine the forces exerted by each child. Since the forces are given in both magnitude and direction, we will use vector addition.
First, let's convert the force exerted by the second child from a magnitude and direction to its horizontal and vertical components. We can use trigonometry to do this.
The force exerted by the second child is 60 N at a direction of southwest (SW). Since southwest is at a 45-degree angle from north (assuming north is the positive y-direction), we can use the following equations:
horizontal component = magnitude * cos(angle)
vertical component = magnitude * sin(angle)
horizontal component = 60 N * cos(45°) = 60 N * 0.707 = 42.42 N [E]
vertical component = 60 N * sin(45°) = 60 N * 0.707 = 42.42 N [S]
Now we have converted the force exerted by the second child into its horizontal and vertical components: 42.42 N [E] and 42.42 N [S].
To find the net force, we need to add the horizontal and vertical components of both children's forces separately.
Horizontal component of net force = 45 N + 42.42 N [E] = 87.42 N [E]
Vertical component of net force = 0 N + 42.42 N [S] = 42.42 N [S]
Finally, we need to combine the horizontal and vertical components to find the net force. We can use the Pythagorean theorem:
net force = sqrt((horizontal component)^2 + (vertical component)^2)
net force = sqrt((87.42 N)^2 + (42.42 N)^2)
net force = sqrt(7621.0564 N^2 + 1796.1764 N^2)
net force = sqrt(9417.2328 N^2)
net force = 97.03 N
Therefore, the net force acting on the tube is approximately 97.03 N.