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
See previous post.
To find the net force acting on the tube, we need to combine the forces exerted by the two children. Forces can be represented by vectors, which include both magnitude (strength) and direction.
Given that one child exerts a force of 45 N in the north direction (N) and the other child exerts a force of 60 N in the southwest direction (SW), we need to break down the forces into horizontal and vertical components.
To find the horizontal component of the southwest force, we need to determine the angle between the southwest direction and the horizontal direction. Let's assume that angle to be 45 degrees. Now we can find the horizontal component of the southwest force.
Horizontal component of 60 N[SW] = 60 N * cos(45°)
Using the cosine of 45° (which is equal to √2/2), we can calculate:
Horizontal component of 60 N[SW] = 60 N * (√2/2) = 30√2 N
The net force acting on the tube can be found by adding the horizontal components of the forces exerted by both children. Since the forces are in opposite directions, subtract the horizontal component of the north force from the horizontal component of the southwest force:
Net force = Horizontal component of 60 N[SW] - Horizontal component of 45 N[N]
Net force = 30√2 N - 45 N
To simplify further, we can approximate the value of √2 to 1.41:
Net force ≈ 30 * 1.41 N - 45 N
Net force ≈ 42.3 N - 45 N
Net force ≈ -2.7 N
The net force acting on the tube is approximately -2.7 N, pointing in the west direction.