A physics spy, Jim, rescues a physics student from distractions as illutrated be

Question

A physics spy, Jim, rescues a physics student from distractions as illutrated below. The 1200 kg helicopter is pushed upward with a force of 20.0 kN. The spy weighs 800 N, and the mass of the student is 60.0 kg. Find the acceleration of the helicopter and tension in each rope.

Professors Hints: You need 3 Free Body Diagrams and 3 sums of the components of the forces for this problem. Notice that you are given the weight of the spy, but the mass of the student.
The answer to the problem is 5.10 m/s^2,894 N, 2.11kN but I just don't know why

The diagram consists of three verticle circles (the helicopter, spy, and student) with a line between the helicopter and spy and another one between the spy and the student

Explanation / Answer

Upwards force = 20 kN
Downwards force = Helicopter + spy + student
Helicopter weight = 1200 kg * 9.8 = 11760N
Spy weight = 800 N ---> Mass = 800/9.8=81.63kg
Student weight = 60kg*9.8 = 588N

Acceleration: Net force = 20000 - 11760 - 800 - 588 = 6582N
F = m*a = 6852 = (1200 + 60 + 81.63)a
a=5.1 m/s/s

Tension

Bottom rope:To accelerate at 5.1 m/s2 upwards tension must overcome his weight and then add an extra force to not fall down and then have an extra force for upwards acceleration.

F = T = m*a = 60kg(9.8+5.1) = 894N

Top rope: Same thing, just add the spy's mass also

F = T = m*a = (60+800/9.8)(9.8+5.1) = 2100N

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