Question

You are in a helicopter towing a 134-kg laser detector that is mapping out the thickness of the Brunt Ice Shelf along the coast of Antarctica. The original cable used to suspend the detector was damaged and replaced by a lighter one with a maximum tension rating of 319 pounds, not much more than the weight of the detector. The replacement cable would work without question in the case that the detector and helicopter were not accelerating. However, some acceleration of the helicopter is inevitable. In order to monitor the tension force on the cable to make sure the maximum is not exceeded (and therefore to not lose the very expensive detector) you calculate the maximum angle, the cable can make with the vertical without the cable exceeding the tension limit. (a) Assuming straight and level flight of the helicopter, what is that maximum angle? (b) What is the corresponding acceleration? (c) Your colleague wants to add a 9.00-kg infrared camera to the detector. What is the maximum allowable angle now?

Answer 1

Assuming the direction of the helicopter in +x direction.

The "ma" in -ve x direction is the pseudo force acting on the block. (due to the helicopter's motion in +ve x direction)

a.

Equating,

mg =Tcos

319 pound force = 1418.98 N

(134)(9.8) = (1418.98)cos

cos = 0.92545

So, = cos^-1(0.92545) = 22.26^o

b.

Again equating,

ma = Tsin

So, a = Tsin/m = [(1418.98)(sin 22.26) / 134] m/s² = 4.01 m/s²

c.

After adding 9.00 kg , now total mass = 134+9 kg = 143 kg

Equating,

mg =Tcos

(143)(9.8) = (1418.98)cos

cos = 0.9876

So, = cos^-1(0.9876) = 9.03^o