A helicopter flies to the right (in the +x direction) at a constant speed of 14 m/s, parallel to the surface of the ocean. A 1200 kg package of supplies is suspended below the helicopter by a cable as shown in the figure: the package is also traveling to the right in a straight line, at a constant speed of 14 m/s.
(a) Choose the package as the system. What is the rate of change of the momentum of the system?
(c) Represent the package as a dot. Draw a careful force diagram ("free-body diagram") showing all the forces acting on the package. You should have the same number of arrows as you had objects in your list in part (b), and the tail of each arrow should be on the package. Label each force with the name of the object exerting the force.
(d) What is the magnitude of the tension in the cable supporting the package? Carefully show all steps in your work.
|vector Fcable| =
N
(e) Write the force exerted on the package by the cable as a vector.
vector Fcable =
N
(f) What is the magnitude of the force exerted by the air on the package?
|vector Fair| =
N
(g) Write the force on the package by the air as a vector.
vector Fair =
N
(h)
The pilot is concerned about whether or not the cable, whose breaking strength is listed at 12400 N, is strong enough to support this package under these circumstances. Will the cable break?
no
yes
(a) The rate of change of momentum of the system is zero. Since both the helicopter and the package are moving with the same velocity in the same direction, there is no change in momentum.
(c) The force diagram for the package would include the following forces:
- The gravitational force acting downwards, exerted by the Earth.
- The tension force in the cable, exerted by the helicopter.
- The air resistance force, exerted by the air.
(d) To find the magnitude of the tension in the cable, we need to consider the vertical forces acting on the package. Since the package is not accelerating vertically, the sum of the vertical forces must be zero. The tension force in the cable is equal in magnitude and opposite in direction to the weight of the package, which is given by the equation:
Tension = Weight = mass * gravity
Tension = 1200 kg * 9.8 m/s^2 = 11760 N
Therefore, the magnitude of the tension in the cable is 11760 N.
(e) The force exerted on the package by the cable can be represented as a vector pointing upwards, opposite in direction to the gravitational force. It has a magnitude of 11760 N.
(f) The magnitude of the force exerted by the air on the package depends on factors such as air density, the shape of the package, etc. Without additional information, we cannot determine its exact value.
(g) The force on the package by the air can be represented as a vector opposite to the direction of motion of the package. It is typically called the air resistance force.
(h) No, the cable will not break under these circumstances. The tension in the cable, 11760 N, is below its breaking strength of 12400 N.
(a) The rate of change of momentum of the system can be calculated by finding the change in momentum over time. Since the speed of the package is constant and there are no external forces acting on it, the rate of change of momentum is zero.
(b) List of objects:
- Helicopter
- Package
(c) Free-body diagram:
In the free-body diagram of the package suspended below the helicopter, the forces acting on it are:
- Weight (mg) downwards
- Tension in the cable (T) upwards
(d) To determine the magnitude of the tension in the cable, we can use Newton's second law (F = ma). In this case, the only force acting on the package is its weight. So we have:
mg - T = ma
Since the package is not accelerating (moving at a constant speed), the acceleration is zero. Thus:
mg - T = 0
Therefore, the tension in the cable is equal to the weight:
|vector Fcable| = mg = 1200 kg * 9.8 m/s^2 = 11760 N
(e) The force exerted on the package by the cable can be written as a vector:
vector Fcable = -T (upwards)
(f) The magnitude of the force exerted by the air on the package is dependent on factors such as air resistance and wind speed. However, since it is not given in the information provided, we cannot determine the exact magnitude.
(g) The force on the package by the air can be written as a vector:
vector Fair = -F (opposite to the direction of motion)
(h) Based on the information provided, we cannot determine if the cable will break or not. We only know the breaking strength of the cable (12400 N), but we would need to compare it to the tension in the cable to make a conclusion.
To calculate the rate of change of momentum of the system, we need to use the equation:
Rate of change of momentum = mass × rate of change of velocity.
In this case, the mass of the package is given as 1200 kg, and the rate of change of velocity (which is the same as the helicopter's speed since the package is traveling at the same speed) is 14 m/s.
So, the rate of change of momentum is:
Rate of change of momentum = 1200 kg × 14 m/s = 16800 kg⋅m/s.
(a) The rate of change of momentum of the system is 16800 kg⋅m/s.
(c) To draw a free-body diagram for the package, we need to consider all the forces acting on it. In this case, there are two forces:
1. Weight (mg): This force is exerted downwards due to gravity and can be calculated by multiplying the mass (m = 1200 kg) of the package by the acceleration due to gravity (g = 9.8 m/s^2). The weight can be represented as an arrow pointing downwards from the package.
2. Tension in the cable: This force is exerted upwards by the cable and counteracts the weight. The tension in the cable is the force we need to find.
(d) To find the magnitude of the tension in the cable, we can use Newton's second law, which states that the net force on an object is equal to mass times acceleration.
Since the package is moving at a constant speed in a straight line, there is no acceleration in the x-direction, so the net force in the x-direction is zero.
Therefore, the tension in the cable must be equal to the weight of the package:
Tension = Weight = mg = 1200 kg × 9.8 m/s^2 = 11760 N.
So, the magnitude of the tension in the cable is 11760 N.
(e) The force exerted on the package by the cable is equal to the tension in the cable. Therefore,
Vector Fcable = Tension = 11760 N.
(f) The magnitude of the force exerted by the air on the package depends on factors such as air resistance, wind, and other atmospheric conditions. Without specific information, we cannot determine the exact magnitude.
(g) The force on the package by the air can be represented as:
Vector Fair = Unknown (since the magnitude is not given).
(h) To determine if the cable will break, we need to compare the tension in the cable with its breaking strength. The breaking strength of the cable is given as 12400 N.
Since the tension in the cable (11760 N) is less than the breaking strength (12400 N), the cable will not break under these circumstances.
So, the answer is "no," the cable will not break.
the answer to part a is that there is no momentum rate of change ( acceleration) because velocity is constant. Thus all forces on the systems must balance to zero in every direction.
The drag is balanced by the horizontal component of cable tension
The weight mg, is balanced by the vertical component of cable tension.
Without knowing the cable angle from horizontal, I can help you no more.