A heavy hatch on a ship is made of a uniform plate of steel that measures 1.2 m X 1.2 m and has a mass of 400 kg. The hatch is hinged along one side; it is horizontal when closed and opens upward. A torsional spring assists in the opening of the hatch. The spring exerts a torque of 2.0 X 10^3 N ■ m when the hatch is horizontal and a torque of 0.3 X 10^3 N • m when the hatch is vertical; in the range of angles between horizontal and vertical, the torque decreases linearly with the angle (e.g., the torque is 1.15 X 10^3 N • m when the hatch is at 45°). At what angle will the hatch be in equilibrium so that the spring exactly compensates the torque due to the weight?
the torque due to weight should be:
$$\tau_h = rmgcos\theta$$
My idea was to set up two equations one for torque of the hatch and one for the torque of the spring setting both equations equal and then solve for the angle My problem is trying to find the eq for the torque of the spring given the data
my try was:
$$\tau_s = \frac{\theta-105}{\frac{-90}{1700}}$$
however this is wrong. can some one help me figure out the eq of the torque of the spring given the data and then help me with the separation of varibles to solve fot the angle.
Since the torque of the elastic force, weight force and external force are equal to:
$$ \tau_1(\theta) = \tau_i - (\tau_i - \tau_f)\frac{2\theta}{\pi}, \quad \quad \quad \tau_2(\theta) = m\,g\,\frac{L}{2}\,\cos\theta, \quad \quad \quad \tau_3(\theta) = F\,L $$
with $0 \le \theta \le \frac{\pi}{2}$, the minimum force that a sailor must apply to open the hatch is equal to:
$$ \tau_1(0) + \tau_3(0) = \tau_2(0) \quad \quad \Leftrightarrow \quad \quad F = 295\,N. $$
Thus, the spring torque equals the weight force torque when:
$$ \tau_1(\theta) = \tau_2(\theta) \quad \quad \Leftrightarrow \quad \quad \theta = 1.33\,\text{rad} = 76.2°. $$
Finally, the minimum force that the sailor will have to apply to close the hatch is equal to:
$$ \tau_2(\pi/2) + \tau_3(\pi/2) = \tau_1(\pi/2) \quad \quad \Leftrightarrow \quad \quad F = 250\,N. $$
About the resolution of the second equation, defined the function:
$$ f(\theta) := \tau_1(\theta) - \tau_2(\theta) $$
and chosen a first attempt angle:
$$ \theta_0 = \frac{0+\pi/2}{2} = \frac{\pi}{4} $$
according to the Newton-Raphson method, we have:
$$ \theta_k = \theta_{k-1} - \frac{f(\theta_{k-1})}{f'(\theta_{k-1})} \quad \quad \text{with} \; k = 1,2,\dots $$
iterations that must be stopped, for example, when:
$$ \left|\theta_k-\theta_{k-1}\right| \le 10^{-3}. $$