A 1000kg gondola is operated on a cable between two towers 340m apart. When the gondola is exactly between the two towers it is 100m below their height. Calculate the tension at the midpoint.

Question

Vocaloid · Answer

Please do not comment unless you are making a serious attempt at a solution.

Narad · Answer

Let the tension in the cables be T 
By symmetry, the tension is the same in both parts of the cable.
The mass = 1000 kg
When the system is in equilibrium
$$\sum_{}^{} Forces =0$$
Therefore,
$$2Tcos \theta =Mg$$
By resolving in the vertical direction
where the angle betweeen the cable and the vertical is
$$\theta = \tan ^{-1} (170/100)$$
$$g=9.81 ms ^{-2}$$
$$T=(Mg)/\cos \theta $$
$$=1000*9.81/\cos \theta =1000*9.81/0.507= 19348N$$

Vocaloid · Answer

that's what I thought too at first but the answer key marks the solution as 10k N :/

Vocaloid · Answer

@sillybilly123 would you mind taking a look at this when you get a chance? Narad and I both get ~20k Newtons but the textbook solution is 10k N

Vocaloid · Answer

(also this is an MCAT problem where we don't have calculators so they just expect us to use 10 as g instead of 9.81 lol)

Gdeinward · Answer

I know this may not be of help, but considering my education level here is a site that features the exact problem: https://forums.studentdoctor.net/threads/another-physics-question.1076530/

Gdeinward · Answer

it seems to do a great job of explaining it.

Narad · Answer

The mistake is
$$2Tcos \theta =Mg$$
$$\cos \theta =0.5$$
$$M=10000 N$$
$$T=(Mg)/(2\cos \theta )$$
$$=10000/(2*0.5)=10000 N$$

sillybilly123 · Answer

always draw the FBD, even if it's obvious.  Cough, splutter