Question

You are in a helicopter towing a 125-kg laser detector that 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 inevitableIn 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 Assuming straight and level of the helicopter, what is that maximum angle? () What is the corresponding acceleration? ( c) Your colleague wants to add a 10.0 kg infrared camera to the detector. What is the maximum allowable angle now?

Answer 1

Answer:

(a) $\theta=25.45^o$

(b) $a=4.665\ m/s^2$

(c) $\theta'=12.80^o$

Explanation:

Accelerated Motion

The kinematic conditions of an object that is subject to forces are mathematically described by Newton's laws, specifically, the second law where the net force Fn can be calculated if we know the mass of the object m and its acceleration a by the relation

$F_n=m.a$

The net force can be also calculated by adding all the vector forces applied to the object. This procedure can be used to compute the acceleration of the object, given the forces acting on it.

In our problem, we have a m=125 Kg laser detector that is being pulled out by a cable of maximum tension T=305 lb-f from a helicopter. As the helicopter moves, the cable makes a vertical angle $\theta$ that must be kept at some limit when the helicopter accelerates in such a way the cable won't break.

The free-body diagram is shown in the figure below. The equilibrium equation for the y-axis is:

$Tcos\theta-W=0$

Rearranging

$Tcos\theta=W$

We assume there is no acceleration in the vertical direction

Similarly, the equilibrium equation for the x-axis is

$Tsin\theta=m.a$

(b) To find the acceleration we must eliminate the angle from the system of equations, we do that by squaring both and adding them term by term:

$(Tcos\theta)^2+(Tsin\theta)^2=W^2+(m.a)^2$

Operating and knowing that W=m.g

$T^2cos^2\theta+T^2sin^2\theta=m^2(g^2+a^2)$

Since $sin^2\theta+cos^2\theta=1$

$T^2=m^2(g^2+a^2)$

Solving for a

$\displaystyle a=\frac{1}{m}\sqrt{T^2-W^2}$

The maximum tension is

$T=305\ lbf=1356.71 Nw$

The Weight is

$W=125 \cdot 9.8=1225\ Nw$

Plugging in the values

$\displaystyle a=\frac{1}{125}\sqrt{1356.71^2-1225^2}$

$a=4.665\ m/s^2$

(a) To find the angle, we divide the second equation by the first equation:

$\displaystyle \frac{Tsin\theta}{Tcos\theta}=\frac{m.a}{m.g}$

Simplifying

$\displaystyle tan\theta=\frac{a}{g}=\frac{4.665}{9.8}=0.476$

Which gives us an angle

$\theta=25.45^o$

(c) If an additional mass of 10 kg was added to the infrared camera, the new weight would be

$W'=(125+10)\cdot 9.8=1323\ Nw$

The new maximum acceleration is now

$\displaystyle a'=\frac{1}{135}\sqrt{1356.71^2-1323^2}$

$a'=2.226\ m/s^2$

With a maximum angle of

$\displaystyle tan\theta'=\frac{2.226}{9.8}=0.227$

$\theta'=12.80^o$