Answer
(a) $T=22.6N$
(b) $T=77.4N$
Work Step by Step
Without upward lift, the guideline slopes downward with angle $\theta$. As shown in the figure below, tension $T$ in the string has 2 components:
- Vertical component $T\sin\theta$ balancing the plane's weight $mg$ $$T\sin\theta=mg=0.9\times9.8=8.82N (1)$$
- Horizontal component $T\cos\theta$ providing the centripetal force $$T\cos\theta=F_c=\frac{mv^2}{r}$$
Because the guideline slopes, the radius $r$ of the rotation circle this time does not equal $17m$, however. Instead, $r=17\cos\theta$
$$T\cos\theta=\frac{0.9v^2}{17\cos\theta}$$ $$T\cos^2\theta=0.053 v^2 (2)$$
Using these two equations, we can find $T$.
(a) For $v=19m/s$, equation (2): $T\cos^2\theta=19.13N$
Dividing (2) over (1), we have $$\frac{T\cos^2\theta}{T\sin\theta}=\frac{1-\sin^2\theta}{\sin\theta}=\frac{19.13}{8.82}=2.17$$ $$1-\sin^2\theta=2.17\sin\theta$$ $$\sin^2\theta+2.17\sin\theta-1=0$$ $$\sin\theta=0.3905$$ $$T=\frac{8.82}{\sin\theta}=22.6N$$
(b) For $v=38m/s$, equation (2): $T\cos^2\theta=76.53N$
We carry out the same steps as in (a) and get $\sin\theta=0.114$
Therefore, $$T=\frac{8.82}{\sin\theta}=77.4N$$
