# Physics question on Newton's Second Law

a 66.5 Kg chandelier is suspended 1.5 metres below a ceiling by three wires. Each wire has the same Tension and the same Length of 2 metres. What is the Tension in each wire?

 Aurora2000 Posts: 111, Reputation: 83 Junior Member #2 Jan 16, 2012, 02:22 AM
The candelier generates a force $F=mg=66.5kg \times 9.81m/s^2$ due to gravity. The 3 wires
must balance this force in order to create an equilibrium. Thus the combined force on the 3 wires is
exactly $F$. As the problem tells you that the tension on each wire is the same, then you have that each wire must have tension $F/3$.
 ebaines Posts: 10,055, Reputation: 5539 Expert #3 Jan 16, 2012, 07:23 AM
Quote:
 Originally Posted by Aurora2000 each wire must have tension $F/3$.
Not quite right. This would be true if all 3 wires were vertical, but they're not. Given that the chandelier is hanging 1.5m below the ceiling, and each wire is 2m long, this tells you that the wires are at an angle $\theta$ to the vertical, where

$cos\theta = \frac {1.5} 2$.

The vertical component of force in each wire must be 1/3 the weight as you show, but the tension in them is greater than that, as there is both a horizontal and vertical component of tension in the wire. So:

$
T = \frac {W/3 } {\cos(\theta) }
$

where W = the weight of the chandelier, in Newtons.
 Aurora2000 Posts: 111, Reputation: 83 Junior Member #4 Jan 16, 2012, 09:29 AM
Yes, ebaines is right, I have not considered wire length.
 seancoakley Posts: 8, Reputation: 10 New Member #5 Jan 16, 2012, 12:32 PM
So, I hope I've got this right.
If the 3 wires were all vertical, the problem would be easier to solve, since all you would have to do would be to work out the weight of the Chandelier, which I calculate to be 652.37 Newtons. Then, in order for equilibrium to occur (i.e the Sum of the forces = 0), then each wire would have a tension of 217.46 Newtons.
However, as ebaines points out, each wire is at an angle to the vertical, which I calculate to be 41.4 degrees.
Thus, the tension in each wire would be greater than 217.46 Newtons. If 217.46 Newtons is the vertical, then:
cos 41.4 = 217.46 divided by T
(i.e. cos theta (41.4) = adjacent (217.46) divided by the hypotenuse (T)).
Therefore, T = 217.46 divided by 0.75 = 289.89 Newtons.
Hope I did this right!!!

## Check out some similar questions!

A train is travelling up a 3.73 degree incline at a speed of 3.25 metres per second, when the last cart breaks free and begins to coast without friction. (i) How long does it take for the last cart to come to rest momentarily (ii) How far did the last cart travel before (momentarily) coming to...

Physics Question on Ideal Gas Law [ 7 Answers ]

A glass bulb of volume 400cm cubed is connected to another of volume 200cm cubed by a tube of negligible volume. Both bulbs initially contain dry air at 20 degrees celsius and 1 atm. The larger bulb is then immersed in steam at 100 degrees celsius and the smaller in melting ice at 0 degrees...

A child on a toboggan (combined weight 70kg) is pulled from rest on a level surface by two friends. The first friend pulls with a force of 100N in the North-West direction and the second friend pulls with a force of 60N in the direction 20 degrees East of North. What is the net force exerted on the...

You observe an arti cial satellite orbiting the earth. You estimates it is at an altitude h = km above the earth's surface and has a mass m = 3500 kg. You wish to calculate when the satellite will be back in the same position. From the second law of motion and the gravitational force law,...

Newton's Second Law [ 2 Answers ]

I need help understanding how to solve this problem F=ma if: how much force is needed to acceleratea 1000-kg car at a rate of 3m/s-squared I don't understand the steps involved. Thanks for your help.