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
Fn = 45N @ 90 Deg. + 60N @ 225 Deg.
X = 45*cos90 + 60*cos225 = -42.43 N.
Y = 45*sin90 + 60*sin225 = 2.57 N.
tanAr = Y/X = 2.57 / -42.43 = -0.06066.
Ar = -3.47 Deg.
A = -3.47 + 180 = 176.5 Deg.
Fn = X/cosA = -42.43 / cos176.5 = 42'51 N. @ 176.5 Deg.,CCW.
To determine the net force acting on the tube, we need to find the vector sum of the forces applied by the two children.
In this case, one child exerts a force of 45 N[N] (north) and the other child exerts a force of 60 N[SW] (southwest).
To find the net force, we need to break down the force of 60 N[SW] into its components. SW stands for southwest, which is a combination of south and west.
Using trigonometry, we can find the vertical (north-south) and horizontal (east-west) components of the force.
Let's assume that the westward component is x and the southward component is y.
Given:
Total force, F = 60 N
Angle between force and west direction, θ = 45 degrees (SW is 45 degrees between south and west)
Using trigonometry, we can determine the vertical and horizontal components:
Vertical Component (y) = F * sin(θ)
= 60 N * sin(45 degrees)
= 60 * 0.7071
= 42.43 N (approximately)
Horizontal Component (x) = F * cos(θ)
= 60 N * cos(45 degrees)
= 60 * 0.7071
= 42.43 N (approximately)
Since the force exerted by the child in the north direction is 45 N and the force exerted in the south direction is 42.43 N, the net vertical force is:
Net vertical force = Force north - Force south
= 45 N - 42.43 N
= 2.57 N (approximately)
Since the horizontal components of both forces cancel each other out (42.43 N in the west direction and 42.43 N in the east direction), the net horizontal force is:
Net horizontal force = Force east - Force west
= 0 N - 0 N
= 0 N
Therefore, the net force acting on the tube is approximately 2.57 N[N].