Three children are fighting over a toy. One child pulls on the toy with a force of 100N at an angle of 37° north of east. The second child pulls on the toy with a force of 200N south. The third child pulls with a force of 250N at an angle of 20° south of west. Watching all of this from nearby, you decide that you have had enough of the bickering, and so you go over and take hold of the toy in order to prevent the children from dragging it (and themselves) away. What equilibrant (magnitude and angle) should you apply to the toy to prevent it, the children, and you from being pulled away? Please help. I really don't understand this problem!
To find the equilibrant force, we need to balance out the forces acting on the toy by applying an equal force in the opposite direction.
Step 1: Break down the given forces into their x and y-components.
- The first child's force (F1) has a magnitude of 100N and an angle of 37° north of east.
- The second child's force (F2) has a magnitude of 200N and acts south.
- The third child's force (F3) has a magnitude of 250N and an angle of 20° south of west.
For the x-component (horizontal direction):
F1x = F1 * cos(angle)
F1x = 100N * cos(37°) = 80.22N (acting east)
F2x = F2 * cos(180°) (180° because south is in the negative x-direction)
F2x = 200N * cos(180°) = -200N (acting west)
F3x = F3 * cos(-20°) (-20° because south-west is between the negative x-axis and the negative y-axis)
F3x = 250N * cos(-20°) = -232.96N (acting west)
For the y-component (vertical direction):
F1y = F1 * sin(angle)
F1y = 100N * sin(37°) = 61.44N (acting north)
F2y = F2 * sin(180°)
F2y = 200N * sin(180°) = 0N
F3y = F3 * sin(-20°)
F3y = 250N * sin(-20°) = -84.88N (acting south)
Step 2: Add up the x and y-components separately to get the resultant force in each direction.
Fx = F1x + F2x + F3x
Fx = 80.22N + (-200N) + (-232.96N) = -352.74N (net force acting west)
Fy = F1y + F2y + F3y
Fy = 61.44N + 0N + (-84.88N) = -23.44N (net force acting south)
Step 3: Find the magnitude and direction of the equilibrant force.
The magnitude of the equilibrant force (Fe) is the magnitude of the resultant force (sqrt(Fx^2 + Fy^2)):
Fe = sqrt((-352.74N)^2 + (-23.44N)^2)
Fe = sqrt(124231.8276N^2 + 548.2336N^2)
Fe = sqrt(124780.0612N^2)
Fe = 352.766 N
The angle (θ) of the equilibrant force is the direction opposite to the resultant force:
θ = atan2(-Fy, -Fx)
θ = atan2(-(-23.44N), -(-352.74N))
θ = atan2(23.44N, 352.74N)
θ ≈ 3.796° south of west
Therefore, to prevent the toy, the children, and yourself from being pulled away, you need to apply an equilibrant force of approximately 352.766N at an angle of 3.796° south of west.
To find the equilibrant force, we need to analyze the forces acting on the toy and determine the net force required to keep it in equilibrium.
Let's first break down the forces into their horizontal and vertical components:
For the first child:
- Force magnitude: 100 N
- Angle: 37° north of east
- Horizontal component: 100 N * cos(37°)
- Vertical component: 100 N * sin(37°)
For the second child:
- Force magnitude: 200 N
- Direction: south (opposite to the positive vertical direction)
- Horizontal component: 0 (no horizontal force)
- Vertical component: -200 N
For the third child:
- Force magnitude: 250 N
- Angle: 20° south of west
- Horizontal component: -250 N * cos(20°)
- Vertical component: -250 N * sin(20°)
Now, let's add up the horizontal and vertical components of the forces:
Horizontal component: 100 N * cos(37°) - 250 N * cos(20°)
Vertical component: 100 N * sin(37°) - 200 N - 250 N * sin(20°)
In order to keep the toy in equilibrium, the net horizontal and vertical forces should be zero. So we set up the following equations:
0 = 100 N * cos(37°) - 250 N * cos(20°) (Equation 1)
0 = 100 N * sin(37°) - 200 N - 250 N * sin(20°) (Equation 2)
Now, we can solve these equations to find the values of the magnitude and angle of the equilibrant force:
Equation 1: 100 N * cos(37°) - 250 N * cos(20°) = 0
Equation 2: 100 N * sin(37°) - 200 N - 250 N * sin(20°) = 0
Solving these equations will give us the values for the magnitude and angle of the equilibrant force you should apply to the toy to prevent it, the children, and you from being pulled away.
Fr = F1 + F2 + F3 + F4 = 0
100N[37o] + 200N[270] + 250N[200] + F4=0
X = 100*Cos37+200*Cos270+250*Cos200 = -155 N.
Y = 100*sin37+200*sin270+250*sin200 =
-225 N.
Tan Ar = Y/X = -225/-155 = 1.45161
Ar = 55.44o = Reference angle.
A = 55.44 + 180 = = 235.4o
F1+F2+F3 = Y/sin A = -225/sin235.4 = 273.3 N.[235.4o] = -155 - 225i.
Fr = -155 - 225i + F4 = 0
F4 = 155 + 225i = 273.3 N[55.44o] = The
equilibrant.
Note: The equilibrant is equal in magnitude but opposite in direction from
the sum of F1,F2, and F3.