Two children on the beach are pulling on an inner tube. One exerts a force of 45 N[N]. The other exerts a force of 60N[SW]. What is the net force acting on the tube
To find the net force acting on the tube, we need to determine the total force resulting from the individual forces applied by each child.
First, we need to understand that forces have both magnitude and direction. The force exerted by the first child is 45 N in the north direction (N), and the force exerted by the second child is 60 N in the southwest direction (SW).
Since forces are vectors, we can break them down into their horizontal (x) and vertical (y) components to facilitate calculations.
To determine the horizontal and vertical components of the second child's force, we can use trigonometry. In this case, the force is acting in the southwest direction, which is a combination of south and west.
To find the horizontal component (Fx) of the second child's force:
Fx = Force x cos(angle)
We know the force is 60 N, but we need to find the angle it makes with respect to the horizontal axis. In this case, the southwest direction is a 45-degree angle with the west direction:
Fx = 60 N x cos(45°)
Using the trigonometric function for cosine, we can calculate:
Fx = 60 N x 0.7071
≈ 42.42 N
To find the vertical component (Fy) of the second child's force:
Fy = Force x sin(angle)
Again, we know the force is 60 N, and the angle is 45 degrees:
Fy = 60 N x sin(45°)
Using the trigonometric function for sine:
Fy = 60 N x 0.7071
≈ 42.42 N
Now we can calculate the net force in the x and y directions by summing up the individual forces:
Net force in the x-direction (horizontal):
45 N (from the first child) + 42.42 N (from the second child) = 87.42 N to the east (since both forces are acting in the same direction)
Net force in the y-direction (vertical):
0 N (since the first child's force is acting only horizontally) + 42.42 N (from the second child) = 42.42 N in the south direction
Finally, we can use vector addition to determine the magnitude and direction of the net force:
Magnitude: √(Net force in x-direction)^2 + (Net force in y-direction)^2
= √(87.42 N)^2 + (42.42 N)^2
≈ √(7622.56 N^2 + 1799.2164 N^2)
≈ √9421.7764 N^2
≈ 97.06 N
Direction: tan^(-1)(Net force in y-direction / Net force in x-direction)
= tan^(-1)(42.42 N / 87.42 N)
≈ 26.25° south of east
Therefore, the net force acting on the tube is approximately 97.06 N in a direction of 26.25° south of east.
It is the vector sum of the two forcesw. Note that they are 135 degrees apart.
You can either add x- and y- components (separately) or use the law of cosines.