Applying Newton's Laws — Question 8

Question 8

A hanging bob can be stationary in an accelerating vehicle

Take $g=10$ m/s$^2$. A 2 kg bob hangs from a light cord inside a vehicle accelerating horizontally right at 5 m/s$^2$. The bob stays at a constant angle relative to the vehicle and shares its acceleration. Ignore air.

1. Find the cord angle from vertical and the side to which the bob trails.

2. Find tension.

3. Why must the force sum not be set to zero merely because the bob is stationary relative to the vehicle?

Question 8 - Solution

A hanging bob can be stationary in an accelerating vehicle

(a) Resolve toward the support.

$$T\sin\theta=ma_x=10,\quad T\cos\theta=mg=20\ \Rightarrow\ \boxed{\theta=\tan^{-1}(1/2)=26.565^\circ}$$ For tilt $\theta$ from vertical, $T\sin\theta=ma_x=10$ N and $T\cos\theta=mg=20$ N. Thus $\tan\theta=1/2$, so $\theta=26.565^\circ$. The bob trails left of its support; tension points upward and right.

(b) Combine perpendicular components.

$$T=m\sqrt{g^2+a_x^2}=2\sqrt{100+25}=10\sqrt5=22.361\,\mathrm N$$ $T=\sqrt{10^2+20^2}=10\sqrt5\approx22.361$ N.

(c) Use an inertial acceleration.

$$\vec F_{\mathrm{physical}}=(10,0)\,\mathrm N=m(5,0),\quad\vec F_{\mathrm{physical}}-m\vec a_{\mathrm{vehicle}}=\vec0$$ The bob has ground acceleration 5 m/s$^2$ rightward despite being at rest in the accelerating vehicle. Ground-frame net force is 10 N rightward. A vehicle-frame equilibrium calculation needs an explicit frame-acceleration term.