Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 93.3 N, Jill pulls with 73.1 N in the northeast direction, and Jane pulls to the southeast with 181 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.
F(x)=93.3 +73.1cos45 +181 cos 45 =
F(y)=73.1 sin45 -182sin 45 =
F=sqrt{F(x)²+F(y)²}=
piuiu
To find the magnitude of the net force exerted on the donkey, we need to add the individual forces together using vector addition.
First, let's represent the given forces graphically using a coordinate system. The east direction will be the positive x-axis, and the north direction will be the positive y-axis.
Jack's force of 93.3 N can be represented as (93.3 N, 0°) since it acts entirely in the east direction (x-axis).
Jill's force of 73.1 N in the northeast direction can be split into two components. The horizontal component can be found by multiplying the force magnitude by the cosine of the angle between the force and the positive x-axis. The vertical component can be found by multiplying the force magnitude by the sine of the angle. The angle between the northeast direction and the x-axis is 45°. Therefore, the horizontal component is 73.1 N * cos(45°) = 73.1 N * 0.707 ≈ 51.7 N, and the vertical component is 73.1 N * sin(45°) = 73.1 N * 0.707 ≈ 51.7 N.
Jane's force of 181 N in the southeast direction can also be split into horizontal and vertical components. Since the southeast direction is the opposite of the northeast direction, we can use the same values as Jill's force. Therefore, the horizontal component is 51.7 N, and the vertical component is -51.7 N (since it points in the opposite direction).
Now, we can add the individual components of the forces together:
Horizontal component: 93.3 N + 51.7 N + 51.7 N = 196.7 N
Vertical component: 0 N + 51.7 N - 51.7 N = 0 N
The net force acting on the donkey can be represented as (196.7 N, 0°) since it only has a horizontal component and no vertical component.
Finally, we can calculate the magnitude of the net force using the Pythagorean theorem:
Magnitude of the net force = sqrt((196.7 N)^2 + (0 N)^2) ≈ 196.7 N
Therefore, the magnitude of the net force exerted on the donkey is approximately 196.7 N.
To find the magnitude of the net force exerted on the donkey, we need to determine the resultant force by combining the individual forces. The magnitude of the net force can be calculated using vector addition.
First, let's represent the given forces as vectors:
Jack's force, F1 = 93.3 N (eastward)
Jill's force, F2 = 73.1 N (northeast)
Jane's force, F3 = 181 N (southeast)
To calculate the net force, we need to add these three vectors together. However, the vectors are not in the same direction, so we need to break them down into their horizontal and vertical components.
Let's break down the forces:
Jack's force in the eastward direction (x-component):
F1x = 93.3 N * cos(0°) = 93.3 N
Jill's force in the northeast direction (x-component):
F2x = 73.1 N * cos(45°) = 51.84 N
Jill's force in the northeast direction (y-component):
F2y = 73.1 N * sin(45°) = 51.84 N
Jane's force in the southeast direction (x-component):
F3x = 181 N * cos(135°) = -127.98 N
Jane's force in the southeast direction (y-component):
F3y = 181 N * sin(135°) = 127.98 N
Now that we have the components, we can add them together to find the resultant force:
Net force in the x-direction:
Fx = F1x + F2x + F3x = 93.3 N + 51.84 N + (-127.98 N) = 17.16 N
Net force in the y-direction:
Fy = F2y + F3y = 51.84 N + 127.98 N = 179.82 N
Now, we can calculate the magnitude of the net force using the Pythagorean theorem:
Magnitude of the net force:
|Fnet| = √(Fx² + Fy²)
|Fnet| = √((17.16 N)² + (179.82 N)²)
|Fnet| ≈ √(294.9456 N² + 32323.4724 N²)
|Fnet| ≈ √(32618.418 N²)
|Fnet| ≈ 180.7 N
Therefore, the magnitude of the net force exerted on the donkey is approximately 180.7 N.