Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 67.3 N, Jill pulls with 98.7 N in the northeast direction, and Jane pulls to the southeast with 133 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.
Fx=67.3+98.7cos45+133cos315=231.1N Fy=0+98.7sin45+133sin315=-24.25N R=sqrt(231.1^2+24.25^2)=232.4N
To find the magnitude of the net force exerted by the people on the donkey, we need to calculate the vector sum of the forces.
First, let's break down the given forces into their horizontal (x) and vertical (y) components.
Jack's force:
- Horizontal component (x): 67.3 N eastward (positive x-direction)
- Vertical component (y): 0 N (no vertical force)
Jill's force:
- Horizontal component (x): 98.7 N * cos(45°) (northeast angle) ≈ 69.9 N in the positive x-direction
- Vertical component (y): 98.7 N * sin(45°) (northeast angle) ≈ 69.9 N in the positive y-direction
Jane's force:
- Horizontal component (x): 133 N * cos(-45°) (southeast angle) ≈ 94.1 N in the positive x-direction
- Vertical component (y): 133 N * sin(-45°) (southeast angle) ≈ -94.1 N in the negative y-direction
Now, let's add up the x-components and y-components separately to find the total x-component (F_netx) and total y-component (F_nety) of the net force.
F_netx = 67.3 N + 69.9 N + 94.1 N
F_nety = 69.9 N - 94.1 N
Next, we can use the Pythagorean theorem to find the magnitude of the net force (F_net) using F_netx and F_nety:
F_net = √(F_netx² + F_nety²)
Substituting the values:
F_net = √((67.3 N + 69.9 N + 94.1 N)² + (69.9 N - 94.1 N)²)
Calculating the expression:
F_net = √((231.3 N)² + (-24.2 N)²)
F_net ≈ √(53494.69 N² + 585.64 N²)
F_net ≈ √54080.33 N²
F_net ≈ 232.6 N
Therefore, the magnitude of the net force exerted by the people on the donkey is approximately 232.6 N.