Question

You are in a helicopter towing a 125-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 305 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 6.00-kg infrared camera to the detector. What is the maximum allowable angle now?

Answer 1

a)

T = tension force in the cable = 305 pounds = 1356.71 N

m = mass of the detector = 125 kg

= maximum angle of the cable with the vertical

Using equilibrium of force in vertical direction

T Cos = mg

(1356.71) Cos = (125) (9.8)

Cos = 1225/1356.71

= Cos^-1(1225/1356.71)

= 25.5 deg

b)

a = acceleration

Along the horizontal direction , force equation is given using Newton's second law as

T Sin = ma

(1356.71) Sin25.5 = (125) a

a = 4.7 m/s²

c)

T = tension force in the cable = 305 pounds = 1356.71 N

m = mass of the detector and camera = 125 + 6 = 131 kg

= maximum angle of the cable with the vertical

Using equilibrium of force in vertical direction

T Cos = mg

(1356.71) Cos = (131) (9.8)

Cos = 1283.8/1356.71

= Cos^-1(1283.8/1356.71)

= 18.87 deg