A rescue helicopter is lifting a man (weight = 868 N) from a capsized boat by means of a cable and harness. (a) What is the tension in the cable when the man is given an initial upward acceleration of 1.83 m/s2? (b) What is the tension during the remainder of the rescue when he is pulled upward at a constant velocity?
I found the first tension= 1030.90N, however I am stuck on trying to figure out the other one. Please help!
(a)
Total tension
= tension due to mass + tension due to acceleration
= mg + ma
(b)
tension at uniform velocity
= tension due to mass only
= mg
To find the tension during the remainder of the rescue when the man is pulled upward at a constant velocity, we can use the concept of equilibrium.
When the man is pulled upward at a constant velocity, the forces acting on him are balanced. This means that the upward force exerted by the rescue helicopter (tension in the cable) is equal to the downward force of gravity acting on the man.
In this case, the man's weight is still 868 N. Since he is being lifted at a constant velocity, we know that the acceleration is zero. That means the net force acting on him is zero.
Using Newton's second law, we can set up the equation:
Sum of forces = mass * acceleration
Since the man is not accelerating, the sum of forces is equal to zero:
Tension - Weight = 0
Now we can substitute the values:
Tension - 868 N = 0
To find the tension, we can solve the equation:
Tension = 868 N
Therefore, the tension in the cable during the remainder of the rescue, when the man is pulled upward at a constant velocity, is 868 N.
It is important to note that the tension in the cable is the same both when the man is given an initial upward acceleration and when he is pulled upward at a constant velocity, as long as the velocity remains constant. Once the man starts moving at a constant velocity, the net force acting on him becomes zero, and therefore the tension equals his weight.