A 27-kg chandelier hangs from a ceiling on a vertical 4.0-m-long wire.
- What horizontal force would be necessary to displace its position 0.15 m to one side?
- What will be the tension in the wire?
In order to watch this solution you need to have a subscription.
This is Giancoli Answers with Mr. Dychko. The chandelier hangs from a rope that's four meters long and it's displaced to the side by 0.15 meters by this horizontal force. And down here I've drawn a free body diagram showing all three forces acting on it, there's gravity straight down there's this tension force along the wire that's on an angle and then there's a horizontal force directly to the right to say. And we can figure out this angle theta, it'll be the same as this angle theta in here because of, you know, interior opposite angles are equal for parallel lines. This vertical component for tension you can imagine as being along here and this and this are parallel, and then these are interior opposite angles this one, and this one that would be in there. Anyway. So, let's calculate theta and then it's inverse sine of the opposite over hypotenuse. So, it's 0.15 meters over 4 meters, and that's 2.149 degrees. And then we can figure out what this tension force is. And then after we know the tension force which will be the answer to part B we'll go tension force multiplied by sine theta which will give us the x component of that tension force has to equal the horizontal force to answer part A. There's other methods to answer this question but I'm going to do part B first. So, we have FTy, the Y component of the tension has to equal gravity. And so, that's ft times cos theta, cos because this is the adjacent leg of this tension force triangle here, and that equals gravity which is mg. And we'll divide both sides by cos theta. And we get tension force is mg over cos theta. So, that's 27 kilograms times 9.8 Newtons per kilogram divided by cos 2.149 degrees. And that gives about 260 Newtons with two significant figures for the tension force. So, the horizontal force has to equal the x component of the tension force, and x component is the opposite leg, and so, we use tension force times sine theta to get it, and that's 264.79 times sine of 2.149 which gives about 9.9 Newtons required to displace that chandelier by 0.15 meters.
if a chandelier experiences a horizontal force, wouldn't the shape be similar to a sector? Instead of a right triangle? B/c I believe that it undergoes pendulum type motion.
This is quite true, but the wording of the question specifies horizontal displacement when it says displace its position 0.15 m to one side. This doesn't mean 0.15 m along the arc it would travel if moved to the side.
Hope this helps,
Mr. Dychko
how is the horizontal force equal to Ftx?
Find us on:
