Two ropes attached to a young tree are pulled by two landscapers. Rope 1 is pulled with 267 N of force at an angle of 23 degrees north of east and Rope 2 is pulled with 377 N of force due west.
a) Determine the magnitude and direction of the resultant force acting on the tree
b) Determine magnitude and direction of a third cable installed to produce equilibrium.
Unless otherwise stated, all angles are measured CCW from +x-axis.
a. F=267[23o] - 377=245.8+104.3i - 377 = -131.2 + 104.3i = 168N[38.5o]N. of W.
b. F3 and F should be equal in magnitude and 180o out of phase:
Therefore, F3 = 168N[38.5o]S. of E.
To solve these problems, we'll use vector addition.
a) To find the magnitude and direction of the resultant force, we need to add the two given forces together.
First, we'll break down the forces into their horizontal (x) and vertical (y) components.
For Rope 1, we have:
Fx1 = 267 N * cos(23°)
Fy1 = 267 N * sin(23°)
For Rope 2, as it is pulled due west (a horizontal force), there is no vertical component:
Fx2 = 377 N
Fy2 = 0
Now, let's add the x-components and y-components separately:
Fx_total = Fx1 + Fx2
Fy_total = Fy1 + Fy2
Substituting the values, we get:
Fx_total = 267 N * cos(23°) + 377 N
Fy_total = 267 N * sin(23°) + 0
Calculating these values:
Fx_total = 267 N * 0.9205 + 377 N ≈ 803.37 N
Fy_total = 267 N * 0.3878 + 0 ≈ 103.47 N
Now, we'll find the magnitude (resultant force) using the Pythagorean theorem:
Resultant Force = √(Fx_total^2 + Fy_total^2)
Substituting the values:
Resultant Force = √(803.37 N^2 + 103.47 N^2)
≈ √(676809.41 N^2 + 10703.68 N^2)
≈ √(687513.09 N^2)
≈ 828.79 N
The magnitude of the resultant force is approximately 828.79 N.
To find the direction, we'll use the inverse tangent (arctan) function:
Direction = arctan(Fy_total / Fx_total)
Substituting the values:
Direction = arctan(103.47 N / 803.37 N)
≈ arctan(0.1287)
Using a calculator or math software, we find:
Direction ≈ 7.39° north of east
Therefore, the magnitude of the resultant force is approximately 828.79 N, and its direction is approximately 7.39° north of east.
b) To determine the magnitude and direction of a third cable needed for equilibrium, we need to find a force that balances out the resultant force.
Since equilibrium means the net force is zero, we can set up the equations:
Fx_resultant + Fx_third_cable = 0
Fy_resultant + Fy_third_cable = 0
Substituting the known values:
Fx_third_cable = - Fx_resultant
Fy_third_cable = - Fy_resultant
Let's calculate these values:
Fx_third_cable = - 803.37 N
Fy_third_cable = - 103.47 N
The magnitude of the third cable's force would then be:
Magnitude of the third cable = √(Fx_third_cable^2 + Fy_third_cable^2)
= √((-803.37 N)^2 + (-103.47 N)^2)
≈ √(645404.94 N^2 + 10701.70 N^2)
≈ √(656106.65 N^2)
≈ 810.12 N
The magnitude of the third cable's force required for equilibrium is approximately 810.12 N.
Since the third cable must balance out the resultant force, its direction would be opposite to that of the resultant force. Hence, the direction of the third cable would be approximately 7.39° south of west.