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In my 100-level university physics course, we are just starting to touch on rigid bodies and tension. While I am fairly certain that I approached this the right way, I would appreciate if someone could look at my work and confirm that this is a valid solution to the following problem:

When you arrive at your favorite restaurant, you are greeted by a large wooden sign. The left end of the sign is held by a bolt, the right end is tied to a rope that makes an angle of $20^\circ$ with the horizontal. If the sign is uniform, $3.2m$ long, and has mass of $16kg$, what is (a) the tension in the rope and (b) the magnitude and direction of force, $\vec P$, exerted by the bolt?

Starting by listing the sums of forces per direction:

$\sum \vec F_x = \vec P_x = \vec T \cos \theta$
$\sum \vec F_y = \vec P_y + \vec T \sin \theta = \vec F_g \rightarrow \vec P_y = \vec F_g - \vec T \sin \theta$
$\sum \tau = \vec P_y (0L) + \vec T \sin \theta (L) = \vec F_g (\frac{1}{2} L) \rightarrow \vec T = \frac{\vec F_g}{2 \sin \theta}$

NOTE: I use $g = 10 m/s^2$ for simplicity.

$\vec T = \frac{\vec F_g}{2 \sin \theta} = \frac{mg}{2 \sin \theta} = \frac{(16 kg)(10 m/s^2)}{2 \sin 20^\circ} = 234 N$
$\vec P_x = \vec T \cos \theta = (234 N) \cos 20^\circ = 220 N$
$\vec P_y = \vec F_g - \vec T \sin \theta = mg - \vec T \sin \theta = (16 kg)(10 m/s^2) -(234 N) \sin 20^\circ = 80 N$
$|\vec P| = \sqrt{(\vec P_x)^2 + (\vec P_y)^2} = \sqrt{(220 N)^2 + (80 N)^2} = 234 N$
$\theta_P = \arctan(\frac{P_y}{P_x}) = \arctan(\frac{80 N}{220 N}) = 20^\circ$

(a) The tension in the rope is $234 N$.
(b) The magnitude of the force exerted by the bolt is $234 N$ while the direction is $20^\circ$ to the horizontal.

I suppose what has me second-guessing myself is that the magnitude and direction of the force exerted by the bolt are the same as the tension, just flipped about the y-axis. Assuming this is correct, is this because the sign is in static equilibrium and the net forces must be zero, so the only way for that to be possible is for forces along the x- and y-axes to balance out, and it just so happens that the tension in the y-direction is half of the force of gravity in this problem?

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  • $\begingroup$ Asking us to check your work is not a good question here. This is not a homework site. $\endgroup$ – sammy gerbil Nov 28 '16 at 0:08
  • $\begingroup$ @sammygerbil And yet, there is a "homework-and-exercises" tag. $\endgroup$ – Jordan Nov 28 '16 at 0:30
  • $\begingroup$ Yes, and to go with it there is a homework (and exercises) policy. The main points are : 1. show your attempt (which you have done), and 2. ask about a conceptual difficulty. Your question in the final paragraph could count as the latter. I apologise, I overlooked that. $\endgroup$ – sammy gerbil Nov 28 '16 at 0:43
  • $\begingroup$ @sammygerbil - if you read all the way to the end of the question, there's a conceptual question here: "is it coincidence that these things balance out, or is that supposed to happen?" $\endgroup$ – Floris Nov 28 '16 at 0:44
  • $\begingroup$ @sammygerbil - we were writing at the same time... and we agree. $\endgroup$ – Floris Nov 28 '16 at 0:47
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Perhaps drawing a couple of diagrams helps answer your question.

enter image description here

The left hand diagram shows the three forces acting on the sign: the attractive force due to the Earth (weight), the force exerted by the rope (tension) and the force on the sign due to the bolt.
Because this is a static equilibrium situation the lines of action of the three forces must all meet at a point which is $X$ in the diagram.

The right hand diagram is the vector addition of the three forces which must be zero as the sign is in static equilibrium.
This is sometime called the "triangle of forces".

The symmetry of the situation shows that the angle that the line of action of the force on the sign due to the bolt relative to the horizontal is $20^\circ$.

I hope that this also shows that this sort of problem can be solved in a few lines using the sine rule.

$\dfrac{16\;g}{\sin 40}= \dfrac{\text{tension}}{\sin 70}= \dfrac{\text{bolt force }}{\sin 70}$.

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  • $\begingroup$ Yay for using diagrams. Funny thing is, I had a different picture in mind (where the sign was between two walls). Yours makes much more sense. Aha moment. $\endgroup$ – Floris Dec 1 '16 at 1:32
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Your statement in the last paragraph is correct. When there is static equilibrium, we know that both the forces and torques must balance.

To balance the forces, we know that the horizontal components of force must be equal and opposite. The sum of the vertical components must equal the force of gravity. And balancing torques means that the two vertical forces must be equal (so the board does not start rotating).

All these things together mean that the two forces must be equal, except for a sign flip in Y.

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