John and Barry are pulling on a fence post with ropes while their friend Pat uses a chain-saw to sheer it at ground-level. John applies a force of 75.0N [40*] on one rope while Barry applies a force of 85.0N [340*] on the other rope. What is the net force on the post? Show your work.
Fn = 75N[40o] + 85N[340o]
X = 75*cos40 + 85*cos340 = 137.3 N.
Y = 75*sin40 + 85*sin340 = 19.14 N.
tan A = Y/X = 19.14/137.3 = 0.13940
A = 7.94o
Fn=X/cosA = 137.3/cos7.94=138.6 N[7.94o]
net force = 139N [7.93 degrees]
To find the net force on the post, we need to find the horizontal components of the forces applied by John and Barry.
The horizontal component of a force can be calculated using the formula:
Horizontal Component = Force * cos(θ)
Where:
- Force is the magnitude of the force.
- θ is the angle the force makes with the horizontal axis.
Let's calculate the horizontal components of the forces applied by John and Barry:
For John:
Horizontal Component of John's Force = 75.0N * cos(40°)
For Barry:
Horizontal Component of Barry's Force = 85.0N * cos(340°)
Note: In trigonometry, angles are measured counterclockwise from the positive x-axis.
Now, let's substitute the values and calculate:
Horizontal Component of John's Force = 75.0N * cos(40°)
Horizontal Component of John's Force ≈ 57.48N
Horizontal Component of Barry's Force = 85.0N * cos(340°)
Horizontal Component of Barry's Force ≈ 84.42N
To find the net force, we need to add the horizontal components of the forces together:
Net Force = Horizontal Component of John's Force + Horizontal Component of Barry's Force
Net Force = 57.48N + 84.42N
Net Force ≈ 141.9N
Therefore, the net force on the post is approximately 141.9N.
To find the net force on the fence post, we need to calculate the vector sum of the forces applied by John and Barry. The net force is determined by both the magnitude and direction of the forces.
First, let's convert the given forces into their vector components. We can use trigonometry to break down each force into its horizontal (x) and vertical (y) components.
John's force:
Magnitude = 75.0N
Direction = 40°
John's horizontal component = 75.0N * cos(40°)
John's vertical component = 75.0N * sin(40°)
Barry's force:
Magnitude = 85.0N
Direction = 340°
Barry's horizontal component = 85.0N * cos(340°)
Barry's vertical component = 85.0N * sin(340°)
Now, we add up the horizontal components and the vertical components separately to find the net force in each direction.
Net horizontal force = John's horizontal component + Barry's horizontal component
Net vertical force = John's vertical component + Barry's vertical component
Finally, we can find the magnitude and direction of the net force using the calculated horizontal and vertical components.
Net force magnitude = sqrt((Net horizontal force)^2 + (Net vertical force)^2)
Net force direction = atan(Net vertical force / Net horizontal force)
Let's calculate the values.
For John's force:
John's horizontal component = 75.0N * cos(40°) ≈ 57.56N (positive x-axis)
John's vertical component = 75.0N * sin(40°) ≈ 48.18N (positive y-axis)
For Barry's force:
Barry's horizontal component = 85.0N * cos(340°) ≈ 75.39N (negative x-axis)
Barry's vertical component = 85.0N * sin(340°) ≈ -57.47N (negative y-axis)
Net horizontal force = John's horizontal component + Barry's horizontal component ≈ 57.56N + 75.39N ≈ 132.95N (positive x-axis)
Net vertical force = John's vertical component + Barry's vertical component ≈ 48.18N - 57.47N ≈ -9.29N (negative y-axis)
Net force magnitude = sqrt((132.95N)^2 + (-9.29N)^2) ≈ √(17690.40N^2 + 86.08N^2) ≈ √17776.48N^2 ≈ 133.36N
Net force direction = atan(-9.29N / 132.95N) ≈ atan(-0.070N) ≈ -4.015° (relative to the positive x-axis)
Therefore, the net force on the fence post is approximately 133.36N at an angle of -4.015° relative to the positive x-axis.