At the beach, 3 children pull on a floating toy. Child 1 plus with a force of 15N [N 24°E], child 2 pulls south and child 3 pulls west. The net force of the toy is zero. Assume that there are no other significant forces acting on the toy.
a) Calculate the magnitude of the forces exerted by child 2 and 3 on the toy.
b) Child 2 lets go and the other children maintain the same force. Calculate the net force on the toy.
a) Well, well, well, it seems like the kids are pulling in all sorts of directions. Let's have some fun with vectors!
To find the magnitude of the forces exerted by child 2 and 3, we need to break down their forces into their respective north and east components.
Child 1 pulls with a force of 15N [N 24°E], which means the east component is 15N * cos(24°) and the north component is 15N * sin(24°).
Child 2 pulls south, so the force in the north direction is 0N, and the force in the east direction is a big fat zero as well.
Child 3 pulls west, which means the east component is 0N, and the north component is another big fat zero.
Since the net force on the toy is zero, the sum of the east components and the sum of the north components should cancel each other out.
So, the magnitude of the force exerted by child 2 on the toy is 0N, and the magnitude of the force exerted by child 3 on the toy is also 0N.
b) Now let's see what happens when child 2 decides to let go. Oh boy, are things going to get wacky!
Since child 2 is no longer pulling, the force exerted by child 2 on the toy is now zero. Child 1 and child 3's forces are still pulling with their original magnitudes.
The east component of the force exerted by child 1 is still 15N * cos(24°), and the north component is still 15N * sin(24°).
Child 3's force is heading west, so the east component is now -15N, and the north component is still 0N.
To find the net force on the toy, we add up the east components and the north components separately.
The sum of the east components is 15N * cos(24°) - 15N, and the sum of the north components is 15N * sin(24°) + 0N.
And that, my friend, is the net force on the toy when child 2 decides to call it quits. Just plug in those numbers into your calculator and let the fun begin!
To solve this problem, we will analyze the forces acting on the toy using vector addition.
a) To find the magnitude of the forces exerted by child 2 and child 3 on the toy, we can use trigonometry.
Given:
Force exerted by child 1 (F1) = 15N [N 24°E]
To find the magnitude of the forces exerted by child 2 (F2) and child 3 (F3), we need to break down the force exerted by child 1 into its x and y components.
F1x = F1 * cosθ
F1y = F1 * sinθ
Where θ = 24° (the angle with respect to the positive x-axis)
F1x = 15N * cos(24°)
F1y = 15N * sin(24°)
F1x ≈ 13.738 N
F1y ≈ 6.153 N
Now, let's consider the net force acting on the toy, which is zero. The forces exerted by child 2 and child 3 must cancel out the force exerted by child 1.
F2x + F3x = -F1x
F2y + F3y = -F1y
Since child 2 pulls south, we can say F2x = 0 and F2y = -F2.
Therefore,
-F2 + F3x = -F1x ----(1)
F3y = -F1y ----(2)
Substituting the calculated values of F1x and F1y into equations (1) and (2), we get:
-F2 + F3x = -13.738 N ----(1)
F3y = -6.153 N ----(2)
Now, let's solve for F2 and F3.
From equation (2):
F3y = -6.153 N
Therefore, F3 = 6.153 N since F3 is in the y-direction only.
From equation (1):
-F2 + F3x = -13.738 N
F3x = 13.738 N
Therefore, F2 = -13.738 N + 13.738 N = 0 N since F2 is in the x-direction only.
The magnitude of the forces exerted by child 2 (F2) and child 3 (F3) on the toy are:
F2 = 0 N
F3 = 6.153 N
b) When child 2 lets go, the force exerted by child 1 and child 3 will still be acting on the toy.
Since there is no force in the x-direction (F2x = 0), the net force in the x-direction will be the same as before, which is -13.738 N.
Since there is no force in the y-direction (F2y = 0), the net force in the y-direction will be the same as before, which is -6.153 N.
Therefore, the net force on the toy, after child 2 lets go, is approximately "-13.738 N [W] + -6.153 N [S]".
To solve this problem, we need to break down the forces into their respective components and analyze them separately.
a) Let's start by breaking down the forces exerted by child 1, child 2, and child 3 into their respective horizontal (x-axis) and vertical (y-axis) components.
1. Child 1's force: 15N [N 24°E]
- The given force can be divided into its x and y-components using trigonometry.
- The x-component will be 15N * cos(24°) and the y-component will be 15N * sin(24°).
2. Child 2's force: Pulling south
- Since child 2 is pulling south, the force will only have a y-component.
- We can represent this force as F2y.
3. Child 3's force: Pulling west
- Similarly, child 3 is pulling west, so the force will only have an x-component.
- We can represent this force as F3x.
Now, since the net force is zero, the sum of the x-components and the sum of the y-components of the forces must be equal and opposite.
Sum of x-components: F3x = - F1x
Sum of y-components: F2y = - F1y
Let's calculate the magnitude of the forces exerted by child 2 and child 3.
To calculate F2y, we need to know the magnitude of F1y. However, that information is missing from the problem statement. Could you provide the magnitude of F1y?
b) If child 2 lets go and the other children maintain the same force, we can now calculate the net force on the toy.
Since child 2's force (F2y) was directed south and had no x-component, it does not affect the net force in the x-direction.
Therefore, the net force on the toy would be equal to the remaining forces in the x-direction:
Net Force in x-direction = F1x + F3x
F1 = 15N[66o],CCW.
F2 = Pull South.
F3 = Pull West.
a. X = 15*cos66 + F3 = 0
F3 = -15*cos66 = -6.10 N.
Y = 15*sin66 + F2 = 0
F2 = -15*sin66 = -13.7 N.
b. X = 15*cos66-6.10 = 0
Y = 15*sin66 = 13.7 N.=13.7N
Fn = X + Yi = 0 + Yi = 13.7i=13.7N[90o]