Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 90.7 N, Jill pulls with 89.1 N in the northeast direction, and Jane pulls to the southeast with 175 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.
Well, it seems like the donkey is really caught between a rock and a hard place or in this case, between Jack, Jill, and Jane! Let's figure out the net force and its direction.
To find the net force, we need to break down the forces into their horizontal and vertical components. Let's do some math magic!
Jack's force, 90.7 N, is purely in the eastward direction. So, its horizontal component is 90.7 N and its vertical component is 0 N.
Jill's force, 89.1 N, is at a northeast angle. To find its horizontal and vertical components, we need to do a little trigonometry. If we split it into its components, we'll get a horizontal component of 89.1 N * cos(45°) and a vertical component of 89.1 N * sin(45°).
Jane's force, 175 N, is at a southeast angle. Again, we'll use some trigonometry to find its components. The horizontal component is 175 N * cos(45°) and the vertical component is 175 N * sin(45°) as well.
Now, let's add up the horizontal and vertical components separately to find the net force:
Horizontal Component: 90.7 N + 89.1 N * cos(45°) + 175 N * cos(45°)
Vertical Component: 0 N + 89.1 N * sin(45°) - 175 N * sin(45°)
Once we have the horizontal and vertical components, we can find the magnitude of the net force using good old Pythagoras:
Magnitude of Net Force = √(Horizontal Component^2 + Vertical Component^2)
Now, for the direction of the net force, we can use some more trigonometry. The angle from the east direction is given by:
Angle = arctan(Vertical Component / Horizontal Component)
So, there you have it! Plug in the values into the equations, and you'll get the magnitude of the net force and its direction. Just remember, the donkey may be stubborn, but don't let those forces pull you in different directions!
To find the magnitude of the net force, we first need to break down the given forces into their horizontal and vertical components.
Given:
Jack's force, F₁ = 90.7 N (eastward)
Jill's force, F₂ = 89.1 N (northeast direction)
Jane's force, F₃ = 175 N (southeast direction)
We can resolve F₂ and F₃ into their horizontal and vertical components using trigonometry.
For F₂:
Horizontal component of F₂ = F₂ * cos(45°) = 89.1 N * cos(45°) = 63.21 N (eastward)
Vertical component of F₂ = F₂ * sin(45°) = 89.1 N * sin(45°) = 63.21 N (northward)
For F₃:
Horizontal component of F₃ = F₃ * cos(45°) = 175 N * cos(45°) = 123.71 N (eastward)
Vertical component of F₃ = F₃ * sin(45°) = 175 N * sin(45°) = 123.71 N (southward)
Now let's add up the horizontal and vertical components separately:
Horizontal component sum = 90.7 N + 63.21 N + 123.71 N = 277.62 N (eastward)
Vertical component sum = 63.21 N (northward) - 123.71 N (southward) = -60.5 N (southward)
The magnitude of the net force can be found using the Pythagorean theorem:
Net force = √[(horizontal component sum)² + (vertical component sum)²]
Net force = √[(277.62 N)² + (-60.5 N)²]
Net force = √[77037.84 N² + 3660.25 N²]
Net force = √80698.09 N²
Net force ≈ 284.15 N
Therefore, the magnitude of the net force the people exert on the donkey is approximately 284.15 N.
To find the direction of the net force, we can use the inverse tangent function:
Direction of the net force = arctan(vertical component sum / horizontal component sum)
Direction of the net force = arctan(-60.5 N / 277.62 N)
Direction of the net force ≈ -12.522°
The direction of the net force is approximately -12.522° from the east direction.
To find the magnitude of the net force, we need to add up the individual forces acting on the donkey.
First, we need to break down the given forces into their horizontal and vertical components.
Jack's force of 90.7 N is purely in the eastward direction (horizontal component), so we can say his force is 90.7 N in the x-direction.
Jill's force of 89.1 N is in the northeast direction. To find its horizontal and vertical components, we can split it into its x and y components. Considering a right-angled triangle, we can use trigonometry.
The angle between the northeast direction and the east direction is 45 degrees, which means that Jill's force can be split into two equal components - one in the east direction and one in the north direction. Using the trigonometric ratios for a 45-degree angle (sin 45° = cos 45° = √2 / 2):
Horizontal component of Jill's force = cos 45° * 89.1 N = (√2 / 2) * 89.1 N ≈ 62.97 N in the x-direction.
Vertical component of Jill's force = sin 45° * 89.1 N = (√2 / 2) * 89.1 N ≈ 62.97 N in the y-direction.
Jane's force of 175 N is in the southeast direction. Again, using trigonometry, we can split it into its x and y components.
The angle between the southeast direction and the east direction is also 45 degrees, which means that Jane's force can be split into two equal components - one in the east direction and one in the south direction.
Horizontal component of Jane's force = cos 45° * 175 N = (√2 / 2) * 175 N ≈ 123.61 N in the x-direction.
Vertical component of Jane's force = -sin 45° * 175 N = - (√2 / 2) * 175 N ≈ -123.61 N in the y-direction.
Now, we can add up the x and y components of all the forces:
Net force in the x-direction = 90.7 N + 62.97 N + 123.61 N ≈ 277.28 N.
Net force in the y-direction = 62.97 N - 123.61 N ≈ -60.64 N.
To find the magnitude of the net force, we calculate the square root of the sum of the squares of the x and y components:
Magnitude of the net force = √(277.28 N)^2 + (-60.64 N)^2 ≈ √76778.40 N^2 ≈ 277.43 N.
So, the magnitude of the net force exerted by the people on the donkey is approximately 277.43 N.
To find the direction of the net force, we can use trigonometry again:
Angle of the net force from the east direction = arctan((-60.64 N) / 277.28 N) ≈ -12.47°.
The negative sign indicates that the angle is south of the east direction.
Therefore, the direction of the net force is approximately -12.47°, indicating that it is South of East.
This is a routine vector addition problem.
Separately add the x (east) and y (north) components of the three force vectors.
Fjack,x= 90.7
Fjack,y = 0
Fjill,x= 89.2 cos45 = 63.07
Fjill.y = 89.2 sin45 = 63.07
Fjane,x = 175cos45 = ______
Fjane,y = -175cos45 = ______
For the resultant,
Fx = Fjack,x + Fjill,x + Fjane,x = ____
Fy = Fjack,y + Fjill,y + Fjane,y = ____
Direction of resultant = arctan Fy/Fx
= ____
Fill in the blanks