# Force on 2 cables with weight attatched using Theta...

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1. Jun 27, 2017

### Unicow

1. The problem statement, all variables and given/known data
https://ibb.co/g6iKDQ

Calculate the magnitude of the force on cables 1 and 2 in the figure below if θ_1 = 55° and θ_2 = 35°.

2. Relevant equations

Is my method incorrect or did I go wrong somewhere in my calculations??

3. The attempt at a solution
So what I did was I used the equation for F_1 + F_2 + F_g = 0.

F_1 = f_1 * <cos(125˚), sin(125˚)>
F_2 = f_2 * <cos(35˚), sin(35˚)>
F_g = 9.8 * (50) = 490<0, -1>

Then set F_1 + F_2 + F_g = <0, 0>

Moving to the system of equations
f_1*cos(125˚) + f_2*cos(35˚) = 0
f_1*sin(125˚) + f_2*sin(35˚) - 490 = 0

Solved for f_2 on first equation
f_2 = -f_1(cos(125˚)/cos(35˚))

plugged it into the 2nd equation
f_1*sin(125˚) + (-f_1(cos(125˚)/cos(35˚))sin(35˚) - 490 = 0

Solved for f_1 and got 502.46 rounded
and for f_2 I got 351.82.

After that, I plugged it into the F_1 and F_2 equations by
F_1 = 502.46<cos(125˚), sin(125˚)>
F_2 = 351.82<cos(35˚), sin(35˚)>

F_1 = <-288.2, 411.59>
F_2 = <288.2, 201.8>

Then got the magnitude with
Sqrt(288.2^2 + 411.59^2) = 502.46
Sqrt(288.2^2 + 201.8^2) = 351.83

Last edited by a moderator: Jun 28, 2017
2. Jun 28, 2017

### BvU

Hello Unicow,

Pictures go with copy/paste in windows. Don't know how to do it on a phone.

The problem statement is somewhat unfortunate: at each end of a cable equal but opposite forces are acting on the cable.
IMHO asking for the tensions in the cable is much clearer. Never mind.

May be correct, but you don't tell us what they are...
Again, you don't tell. I can guess, but that's not the idea!
You do bother to continue finding $\vec F_1, \vec F_2$ (even though the exercise doesn't ask that) but fail to notice that $411.59+201.8-490\ne 0$

And then you go on and calculate the magnitudes you had already !

3. Jun 28, 2017

### Ray Vickson

Whoa! The question said the block weighs 50 lb, so is using the old English system of units That means that your $g$ is wrong; you need to use the value in English units also, or else convert the block's mass to kg.

There is also the issue of whether you should multiply by $g$ at all. In the USA at least, a 50 lb block exerts a force of 50 lb in Earth's sea-level gravity, so the $g$ is already incorporated in the 50 lb figure. (Basically, it is describing a mass by its force and mixing up the two concepts, but in a standard and understood way.)

Last edited: Jun 28, 2017
4. Jun 28, 2017

### Unicow

Yeah, I ended up figuring that out sooner or later haha. Stupid mistake leading to almost an hour of checking over my work mostly. The answer was so simple. Thank you though haha. I tried to incorporate what we used in class but failed to recognize that we used kg in class vs lb.