Calculating Angle of Globe on Accelerating Train - Forces in Motion

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To determine the angle of a hanging globe in an accelerating train, consider the forces acting on it: the gravitational force downward and the apparent force due to the train's acceleration. When the train accelerates at 1.5 m/s², an apparent horizontal force equal to -ma acts on the globe. Using trigonometry and the Pythagorean theorem, the relationship between the forces can be established to find the angle with the vertical. The length of the rope also influences the final angle calculation.
futb0l
Got this out of my book .. had no idea how to do it.
Here's the exact question...

An old light globe hangs by a wire from the
roof of a train. What angle does the globe
make with the vertical when the train is accelerating
at 1.5 m s−2?

Thanks guys.
 
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When the train is accelerating with accelration a, it's no longer an inertial frame.
It looks like all objects in this frame are subjected to an apparent force F=-ma.

So just act like there's a horizontal force -ma acting on the globe.
 
Simple trig. What's the force and direction of gravity? What's the force and direction of the train's acceleration? Apply the Pythagorean theorem to get the magnitude and look at your trig relationships (SOHCAHTOA).
 
I think the length of the rope is also a factor.
 
Thanks BobG!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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