How to find the angle of a force applied to a point in the interior of a disk so that is at equilibrium?

Question

I'm trying to find a way to this problem using euclidean geometry. Is that possible?.

The problem is as follows:

The figure from below shows a pulley whose has an axis in its center. A wire is rolled to the pulley so that one end has a mass of $4\,kg$ hanging from it. In another place situated in the interior of the disk a pin is attached to a thin rope of negligible weight from where a person can pull the rope. Find the angle $\theta$ so that the pulley is at equilibrium when the force applied to the point where the rope is fixed is equal to $70\,N$. Assume that the acceleration due gravity is ($g=10\,\frac{m}{s^2}$). Hint: You may use the approximation of $\cos 16^{\circ}=\frac{24}{25}$.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&37^{\circ}\\ 2.&53^{\circ}\\ 3.&16^{\circ}\\ 4.&74^{\circ}\\ 5.&45^{\circ}\\ \end{array}$

Does it exist a way to solve this geometrically?. I've tried to describe the approach which I've attempted to solve this problem. This is summarized in the figure from below.

Typically it seems easy to spot how to apply torques in am object when is straight as an arrow or its shape is linear, but how to do this when the object has a circular shape?. Can someone help me with this?.

I've tried all sorts of constructions to find the radius ($r_{1}$) which is indicated in the figure from above. But I couldn't. Does it exist a way to relate it with what it is being asked without the necessity of using vector decomposition?.

I'm not sure if the line which crosses to the force is tangent. Does it exist a way to tell if its placed in the right spot?. What would be the maximum angle so that the tangent is situated in those $40\,cm$?. Is it related with the $45^{\circ}$?.

I think if its less than $45^{\circ}$ the line crosses the force below the horizontal line in the disk but if the angle is greater than those $45^{\circ}$ then it will be over the horizontal line. Can someone help me with that part as well?.

Answer 1

Typically it seems easy to spot how to apply torques in am object when is straight as an arrow or its shape is linear, but how to do this when the object has a circular shape?. Can someone help me with this?.

$\text{Torque} = F_{t}\cdot r$ where $F_t$ is the force tangential to the angle of the circle at point where the force acts.

As a start, I find it hard to understand how this problem should be solved. To reach the equilibrium, you must first compute how much the resulting tangential force should be. to compute the angle from the force on the rope and its tangential force, it is simply required to decompose the vector into a relative tangential and normal component. Either way, i will continue with what I got: The resulting tangential force required to reach an equilibrium can be derived from the following equation: $F_tr_2 = mgR = 2800$

First compute the radius $r_2$ between the rope and the center. The vertical part is already given, and the horizontal part can be deduced from the 45 degree angle and the remaining height using the radius (70-40 = 30). As such, the radius equals 50. Therefore $F_t = 56N$ Additionally, $F_t$ is the tangential part of F, with respect to the angle to the center of the disk: $$F_t = F\sin(\theta + \alpha)$$ $\alpha$ is the angle between the line connecting the center and the connecting point on the rope and the vertical axis. this is equal to $\arctan(30/40)$, but that is irrelevant for the solution. Knowing F=70N, the equation can be solved as follows: $$\sin(\theta + \alpha) = 0.8$$ $$\sin\theta \cos\alpha + \cos\theta \sin\alpha = 0.8$$ $$5\sin\theta \cos\alpha + 5\cos\theta \sin\alpha = 4$$ Since we know $\alpha$ is part of that satisfying 3-4-5 triangle, we can rule $\alpha$ out of the equation: $$4\sin\theta + 3\cos\theta = 4$$ $$(4\sin\theta)^2 = (4 - 3\cos\theta)^2$$ $$16\sin^2\theta = 9\cos^2\theta -24\cos\theta + 16$$ $$16-16\cos^2\theta = 9\cos^2\theta -24\cos\theta + 16$$ $$0 = 25\cos^2\theta -24\cos\theta$$ and from here it can be easily seen that $\cos\theta = 0$ or $\cos\theta = \frac{24}{25}$ So that immediately explains why that hint is given and might solve your question to what is the maximum angle. This angle is namely where $\cos\theta = 0$ thus 90 degrees. Maybe this doesnt answer your question to the fullest, but maybe it helps a bit.

Answer 2

A simple way to look at this problem. The rectangle in the first figure has height $40$ cm and base $30$ cm. So its diagonal, which is $r_1$, is $50$ cm. The angle $\alpha$ between $r_1$ and the vertical line traced from the point where the rope is fixed is $\arctan 30/40\approx 0.64$ rad, corresponding to about $37°$. Note that $\sin \alpha=3/5$ and $\cos \alpha=4/5$.

Now note that the effect of our pulling force on the $4$-Kg mass is proportional to $\sin(\alpha+\theta)$: it is maximal if $\theta=\pi/2-\alpha=90°-37°=53°$ (this is the case in which we pull the rope perpendicularly to $r_1$), and reduces to zero if $\theta=-\alpha=-37°$ (this is the theoretical case in which we pull the rope in a direction coinciding with $r_1$).

So, considering that the effect of the pulling force on the $4$-Kg mass is further reduced by a factor equal to $r_1/r=50/70=5/7$, we have to find $\theta $ such that

$$70\cdot \frac 57\, \sin(\alpha+\theta)=40$$

which gives

$$\sin(\alpha+\theta)=\frac 45$$

To solve it, we remind that

$$\sin(a+b)=\sin a \cos b +\cos a \sin b$$

and then

$$\sin \alpha \cos\theta +\cos \alpha \sin \theta =\frac 45$$

$$\frac 35 \cos\theta + \frac 45 \sin \theta=\frac 45$$

$$3 \cos\theta + 4 \sqrt{1-\cos^2 \theta} = 5$$

Solving the quadratic for $\cos\theta$ we get

$$\cos \theta =\frac{24}{25}$$

and then, using the hint

$$\theta=16°$$