# Equilibrium: max distance traveled; undisturbed tension

1. Nov 15, 2015

### Sociomath

• Missing template due to originally being posted in a different forum.
Hi,

I need some guidance on the following problem, please.

A daredevil attempts to walk the full length of suspended. A 20 kg wrecking ball hangs at the end of this uniform beam of length 4 m and mass 10 kg and is attached to a hinged wall at an angle of 53 degrees. A cable attached to the wall at an angle of 60 degrees is tied to the end of the beam and supports a maximum tension of 525 N.

How far along the beam can a man of 70 kg walk without the cable breaking?
(Apologies for the diagram-sketch.)

$Fgm = (\frac12)(4m)(\sin 37°)+ Fgm(x \cdot 4m)(\sin 37°) - F_{wall}(4m)(\sin 53°)$

$F_w =\displaystyle \frac{29.4889 + 412.8451x}{\sin 53°}$

$F_{net} = F_{wall} - F_t$
$F_{wall}- F_t = 0$
$F_{wall} - 525 N =0$
$F_{wall} = 525 N$

$x = \displaystyle \frac{419.265 - 29.4889}{412.8451} = 0.94412$

Finally, multiply 0.94412 by 4 m:
$0.94412 \cdot 4 = 3.776 m$

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Last edited: Nov 15, 2015
2. Nov 15, 2015

### PeroK

What's the question?

In your solution you haven't said what you're doing. It's just a lot of undefined terms and some numbers.

3. Nov 15, 2015

### Sociomath

Thanks, PeroK.

How far along the beam can a man of weight 70 kg walk without the cable breaking?

$\displaystyle \tau_{pivot} = \tau_{cw} - \tau_{ccw}$
d = 4 m = length of beam

$0 = Fgm(\frac12 \cdot 4m) \sin 37° + Fgm(x \cdot 4m) \sin 37° - F_{wall} (4m) \sin 53°$
$F_{wall} (4m) \sin 53° = (9.8 m/s^2)(10 kg)(\frac12 \cdot 4m) \sin 37° + ((9.8 m/s^2)(70 kg)(x \cdot 4m) \sin 37°$

$F_{wall} \sin 53° = (9.8 m/s^2)(10 kg)(\frac12) \sin 37° + ((9.8 m/s^2)(70 kg)(x) \sin 37°$

$F_{wall} =\displaystyle \frac{29.4889 + 412.8451x}{\sin 53°} ...Eq(1)$

Applying the translation equilibrium condition,
$F_{wall} - F_{T} = F_{net} x$
$F_{wall} - 525 N = 0$
$F_{wall} = 525 N$

Now, substituting $F_{wall} = 525 N$ into Eq(1),

$525 N =\displaystyle \frac{29.4889 + 412.8451x}{\sin 53°}$

$x = \displaystyle \frac{419.265 - 29.4889}{412.8451} = 0.94412$

Finally, multiply 0.94412 by 4 m:
$0.94412 \cdot 4 = 3.776 m$
So a man of 70 kg can walk 3.776 meters along the beam before the cable breaks.

4. Nov 15, 2015

### PeroK

I can guess at some of the terms you're using and I can guess you're taking moments about the pivot. I don't see anything I can recognise as tension in the cable nor the influence of the 20kg mass.

I've no idea what F is.