Determining the Angle based in Tension

I am trying to solve the following problem, but got lost in the process.

I am trying to find the minimum/maximum angle ( θ ) needed to have the system in equilibrium without breaking the ropes which max tension can only support up to 700 lbs. Angle AB is fixed. The equations are as follows:

Eq1: ∑ Fx = TAB cos (25) - TAC cos (θ) = 0

Eq2: ∑ Fy = TAB sin (25) + TAC sin (θ) = 1000

Solving Eq1 for TAC, we have:

Eq3: TAC = TAB cos (25) / cos (θ)

You can conclude that at all times TAC is always greater than TAB. Therefore you can set TAC to 700 lbs in either Eq 1 or Eq2 and try to find θ But that still left me with the unknown TAB. How can I find θ ?

• Angle AB can't be fixed, as the weight will cause both ropes to elongate. You might want to double-check the statement of the question. Another note is that the equation for rotational equilibrium is missing.
– r13
Commented Jun 4, 2021 at 22:52

TL;DR: There is something wrong with the problem definition. The minimum force on the $$T_{AC}$$ is approximately 906 lbs at $$\theta=65\deg$$ (in that case $$T_{AB}= 422[lbs]$$)

NOTE: From (Eq.3) what you really get is that:

$$\frac{T_{AC}}{ T_{AB}} =\frac{ cos (25)} {cos (θ)}$$

which actually can be interpreted, as:

• $$T_{AC}>T_{AB}$$ when $$\theta > 25\deg$$
• $$T_{AC} when $$\theta < 25\deg$$

So basically you need to check first what happens at 25$$\deg$$.

at 25$$\deg$$

At 25$$\deg$$, there is symmetry so:

$$T_{AC} =T_{AB}$$

So from the second equation you can obtain that:

$$∑ Fy = T_{AB}= T_{AC}= \frac{1000}{sin (25)}=2366$$

which means that the force is already exceeding 1000.

So you would need to have an angle $$\theta >25$$.

Because then $$T_{AB}, and if $$T_{AC}= 700 [lbs]$$ then $$T_{AB}$$ will also be $$T_{AB}<700$$.

Solving for $$\theta$$.

Since $$\theta >25\deg$$, $$T_{AC}>T_{AB}$$ and therefore you can substitute in Eq.2, the following form of eq.3 $$T_{AB} =T_{AC}\frac{cos (θ)}{ cos (25)}$$ :

$$T_{AB} sin (25) + T_{AC} sin (θ) = 1000$$

$$T_{AC}\frac{\cos (θ)}{ \cos (25)} sin (25) + T_{AC} \sin (θ) = 1000$$

$$T_{AC}\left(\frac{\cos (θ)}{ \cos (25)} sin (25) + \sin (θ)\right) = 1000$$

$$T_{AC}\left(\cos (θ)\tan(25) + \sin (θ)\right) = 1000$$

$$T_{AC}\left(\cos (θ)\tan(25) + \sin (θ)\right) = 1000$$

From this equation we can solve for $$T_{AC} = 1000\frac{1}{\cos (θ)\tan(25) + \sin (θ)}$$, and we can plot this equation.

The problem is that the minimum value of $$T_{AC}$$ is at approx 900[lbs].

The reason is that the maximum value for the expression $$\cos (θ)\tan(25) + \sin (θ)$$ is for $$\theta= 65\deg$$. So for 65 deg, you can get the minimum $$T_{AC}\approx 906$$. (In that case, $$T_{AB} = 422 [lbs]$$).

So, you can't really protect the AC rope from breaking.

if you only needed $$T_{AB}$$ to be under 700[lbs], then you can achieve that for angles $$\theta \ge 45\deg$$. see graph below

#%%
import numpy as np
import matplotlib.pyplot as plt

# %%
xs = np.linspace(25,85,100)
ts = xs*np.pi/180
Ts = 1000/(np.cos(ts)*np.tan(25*np.pi/180)+ np.sin(ts))
TABs = Ts*np.cos(ts)/np.cos(25*np.pi/180)
# %%
plt.figure(figsize=(10,8))
plt.plot(xs,Ts, label='$$T_{AC}$$')
plt.plot(xs,TABs, label='$$T_{AB}$$')
plt.grid()
plt.legend(fontsize=15)
plt.xlabel('Angle[deg]', fontsize=15)
plt.ylabel('Forces [lbs]', fontsize=15)