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.
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
1 answer
1 answer