Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 72.7 N, Jill pulls with 79.9 N in the northeast direction, and Jane pulls to the southeast with 103 N. (Since the donkey is involved with such uncoordinated people, who can blame it for being stubborn?) Find the magnitude of the net force the people exert on the donkey.
What is the direction of the net force? Express this as the angle from the east direction between 0° and 90°, with a positive sign for north of east and a negative sign for south of east.
F1+F2+F3 = 72.7N[0o] + 79.9N[45o] + 103N[315o].
X = 72.7 + 79.9*Cos45 + 103*Cos315=202 N
Y = 79.9*sin45 + 103*sin315 = -16.33 N.
F^2 = X^2 + Y^2 = 202^2 + (-16.33) = 41,071
F = 203 N. = Resultant force.
Tan A = Y/X = -16.33/202 = -0.08084
A = -4.62o = 4.62o S. of E. = Direction.
To find the magnitude of the net force exerted on the donkey, we need to find the components of each force in the eastward and northward directions.
First, let's break down the force exerted by Jack. Since Jack pulls eastward, the entire force of 72.7 N is in the eastward direction. The component of the force acting in the northward direction is 0 N.
Next, let's break down the force exerted by Jill. Jill pulls in the northeast direction. To find the eastward and northward components, we can use trigonometry. The angle between the northeast direction and the east direction is 45 degrees.
The eastward component of Jill's force is given by:
Cosine(45°) * 79.9 N = 79.9 N / sqrt(2) ≈ 56.56 N
The northward component of Jill's force is given by:
Sine(45°) * 79.9 N = 79.9 N / sqrt(2) ≈ 56.56 N
Now let's break down the force exerted by Jane. Jane pulls in the southeast direction. Again, to find the eastward and southward components, we can use trigonometry. The angle between the southeast direction and the east direction is also 45 degrees.
The eastward component of Jane's force is given by:
Cosine(45°) * 103 N = 103 N / sqrt(2) ≈ 72.92 N
The southward component of Jane's force is given by:
Sine(45°) * 103 N = 103 N / sqrt(2) ≈ 72.92 N
Now, to find the net force, we need to add up the eastward and northward components of each force:
Eastward component = 72.7 N + 56.56 N + 72.92 N ≈ 202.18 N
Northward component = 56.56 N + 72.92 N - 72.92 N = 56.56 N
The magnitude of the net force is given by the Pythagorean theorem:
Magnitude = sqrt(Eastward component^2 + Northward component^2)
Magnitude ≈ sqrt(202.18 N^2 + 56.56 N^2)
Magnitude ≈ sqrt(40898.4684 N^2)
Magnitude ≈ 202.21 N (rounded to two decimal places)
To find the direction of the net force, we can use the tangent function:
Tangent(θ) = Northward component / Eastward component
θ = arctan(Northward component / Eastward component)
θ = arctan(56.56 N / 202.18 N)
θ ≈ arctan(0.2798)
θ ≈ 15.75°
Since the angle is measured from the east direction, and we want it to be between 0° and 90°, the direction of the net force is approximately 15.75° north of east.