There is a two member dog sled team pulling with all their strength in the middle of a snow storm. Dog
A is pulling with a force of 178 N at 15�
and Dog B is pulling with 225 N at −10�
. The sled and gear
weigh 980 N and has a coefficient of friction of 0.3 with the thick snow. The dog sled team is aiming for
wooden shelter straight ahead but are being pushed off course −28�
from the wind towards a crevasse. If
the sled team is traveling with a velocity of 11 m/s what is the speed and direction of the wind?
Whoever wrote this problem assumed that the sled would have a velocity component of 11 m/s in the downwind direction due to the wind.
However a sled (or a ship} is not an airplane or a balloon. if nothing else is pushing it, the wind in fact might not move it at all. The ice or water will exert a force counteracting the wind force which depends on the lift and drag coefficients of the sled as well as the lift and drag characteristics of the sled/ice interface. Think about the sled being on tracks or wheels :)
TKIug
To solve this problem, we need to break down the forces acting on the dog sled team. Let's calculate the net force acting on the sled first using vector addition.
1. Resolve the forces applied by each dog into their horizontal (x) and vertical (y) components.
Dog A:
Force in x-direction (𝐹𝐴𝑥) = 178 N * cos(15°)
= 178 N * 0.966
= 171.588 N
Force in y-direction (𝐹𝐴𝑦) = 178 N * sin(15°)
= 178 N * 0.259
= 46.202 N (upward force)
Dog B:
Force in x-direction (𝐹𝐵𝑥) = 225 N * cos(-10°)
= 225 N * 0.985
= 221.625 N
Force in y-direction (𝐹𝐵𝑦) = 225 N * sin(-10°)
= 225 N * -0.174
= -39.150 N (downward force)
2. Calculate the net force in the x-direction (horizontal) by summing up the forces.
Net force in x-direction (𝐹𝑛𝑒𝑡,𝑥) = 𝐹𝐴𝑥 + 𝐹𝐵𝑥
= 171.588 N + 221.625 N
= 393.213 N
3. Calculate the net force in the y-direction (vertical) by summing up the forces.
Net force in y-direction (𝐹𝑛𝑒𝑡,𝑦) = 𝐹𝐴𝑦 + 𝐹𝐵𝑦
= 46.202 N - 39.150 N
= 7.052 N (upward force)
4. Calculate the net force acting on the sled using the net force components.
Net force (𝐹𝑛𝑒𝑡) = √(𝐹𝑛𝑒𝑡,𝑥)² + (𝐹𝑛𝑒𝑡,𝑦)²
= √(393.213 N)² + (7.052 N)²
≈ √(154,578.387 N² + 49.733 N²)
≈ √(154,628.120 N²)
≈ 393.198 N
5. Calculate the frictional force opposing the motion of the sled.
Frictional force (𝐹𝑓) = Coefficient of friction (𝜇) * Normal force (𝑁)
= 0.3 * 980 N
= 294 N
6. Calculate the wind force pushing the sled off-course.
Wind force (𝐹𝑤) = 𝐹𝑓 + 𝐹𝑛𝑒𝑡
= 294 N + 393.198 N
= 687.198 N
7. Calculate the angle of the wind force (direction) relative to the horizontal.
Angle of the wind (𝜃) = arctan(𝐹𝑛𝑒𝑡,𝑦 / 𝐹𝑛𝑒𝑡,𝑥)
= arctan(7.052 N / 393.213 N)
≈ arctan(0.0179)
≈ 1.0266°
8. Calculate the speed of the wind (magnitude of the wind force).
Speed of the wind = 𝐹𝑤 / mass of the sled team (since acceleration = force / mass)
= 687.198 N / 980 kg (convert from N to kg using acceleration due to gravity)
≈ 0.702 m/s
Therefore, the speed of the wind is approximately 0.702 m/s, blowing at an angle of approximately 1.0266° relative to the horizontal.
To solve this problem, we need to analyze the forces acting on the dog sled team and determine the resultant force.
Let's break down all the forces acting on the sled:
1. Dog A's pulling force (Fa): 178 N at 15° is acting forward.
2. Dog B's pulling force (Fb): 225 N at -10° is acting forward.
3. Frictional force (Ff): The coefficient of friction is 0.3, and the weight of the sled is 980 N. Therefore, the frictional force can be calculated as Ff = coefficient of friction * weight = 0.3 * 980 N.
To determine the resultant force, we need to resolve the forces of dog A and dog B into their horizontal (x) and vertical (y) components:
For Dog A:
Fx(A) = Fa * cos(theta) = 178 N * cos(15°)
Fy(A) = Fa * sin(theta) = 178 N * sin(15°)
For Dog B:
Fx(B) = Fb * cos(theta) = 225 N * cos(-10°)
Fy(B) = Fb * sin(theta) = 225 N * sin(-10°)
Next, we can calculate the horizontal and vertical components of the resultant force:
Fx(resultant) = Fx(A) + Fx(B)
Fy(resultant) = Fy(A) + Fy(B)
Now, let's calculate the components:
Fx(resultant) = (178 N * cos(15°)) + (225 N * cos(-10°))
Fy(resultant) = (178 N * sin(15°)) + (225 N * sin(-10°))
The magnitude of the resultant force (Fr) can be found using the Pythagorean theorem:
Fr = sqrt(Fx(resultant)^2 + Fy(resultant)^2)
Now, let's calculate the magnitude of the resultant force:
Fr = sqrt((178 N * cos(15°) + 225 N * cos(-10°))^2 + (178 N * sin(15°) + 225 N * sin(-10°))^2)
To calculate the direction of the resultant force, we can use trigonometry:
theta(resultant) = arctan(Fy(resultant) / Fx(resultant))
Now, let's calculate the direction of the resultant force:
theta(resultant) = arctan((178 N * sin(15°) + 225 N * sin(-10°)) / (178 N * cos(15°) + 225 N * cos(-10°)))
To find the speed and direction of the wind, we need to consider that the resultant force is equal to the wind force, Fw. The wind force can be expressed as:
Fw = coefficient of friction * weight
Now, let's solve for the speed and direction of the wind:
Speed of the wind = magnitude of the resultant force (Fr) / mass of the sled team
Direction of the wind = theta(resultant) + 180°
Plug in the values into the equations and solve for the answer.