# What to assume in this problem of static equilibrium?

I have a question which I'm going to state in verbatim :

You want to hang a picture at a certain place on a wall, but the only available nails are at points 1 ft. to the left and 2 ft. to the right off the edges of the picture (see the attachment). You attach the strings of the appropriate lengths from these nails to the top corners of the picture, as shown, but the picture will not hang straight unless you add a balancing weight of some kind.
If the picture with its frame weighs 10 lb, what is the least possible balancing weight , and where would you put it?

I have written down the problem the way it is because there are chances that I may have misunderstood it. The attachment is from the problem only ( I think that diagram is the most confusing thing in this whole problem).

Now, I have few doubts regarding this problem. If we consider the situation to be as the way I have done in second attachment ( $$\displaystyle T_1$$ and $$\displaystyle T_2$$ are tension forces due to strings ) . After doing this we are free to choose ( I'm free therefore I getting the problem ,others may be clear) among these assumptions :-

$$\displaystyle \Sigma F_y = 0$$
$$\displaystyle \Sigma F_x = 0$$
$$\displaystyle \Sigma \vec{ \tau} = 0$$

And according to which two assumptions we choose we would get different answers. My logic behind those assumptions are such that if the picture is hanging then it must mean that it's not falling and not even accelerating upwards therefore sum of vertical forces may be zero, but it could also happen that initially the string 2 (any string 1 or 2) may have a greater vertical component and hence the picture has moved a little upwards but this upward movement has caused the string 1 to be stretched and hence it's vertical component too will increase and after some time our picture will be in equilibrium but during this course we have lost information about that $$\displaystyle 1 ft.$$ statement. Similarly, if we take one of those $$\displaystyle x$$ components to be unequal then the picture will move a little to sideways but this would cause the other string to be stretched and hence equilibrating at some position after the picture has been translated horizontally. My reasoning for sum of torques to be zero is that the picture ought to be in rotational equilibrium otherwise we will lose the information about both the nails 9 I mean where would the nails gonna be if our picture were to rotate ).

Thank you any help will be much appreciated.

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#### Cervesa

Using a Cartesian grid (see attached diagram), let the origin be the point where $T_1$ meets the upper left corner of the picture frame.

Line of action for $T_1$ is $y=-x$

Line of action for $T_2$ is $y=\dfrac{x}{2}-1$

Point of intersection for these two lines of action is $\left(\dfrac{2}{3},-\dfrac{2}{3}\right)$

Line of action for the 10 lb weight is $x = 1$

Let the point of intersection for the lines of action for $T_1$ and $T_2$ be the pivot point $P$.

Since $T_1$ and $T_2$ act through point $P$, they exert no torque relative to point $P$.

Relative to point $P$, the 10 lb weight exerts a clockwise torque, $\tau = \dfrac{10}{3} \text{ ft-lbs}$.

To counteract that torque, a weight must be hung that exerts a torque of $\dfrac{10}{3} \text{ ft-lbs}$ in the counterclockwise direction. The least weight would be a 5 lb weight hung on the left edge of the picture frame $\dfrac{2}{3} \text{ ft}$ to the left of point $P$.

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Using a Cartesian grid (see attached diagram), let the origin be the point where $T_1$ meets the upper left corner of the picture frame.

Line of action for $T_1$ is $y=-x$

Line of action for $T_2$ is $y=\dfrac{x}{2}-1$

Point of intersection for these two lines of action is $\left(\dfrac{2}{3},-\dfrac{2}{3}\right)$

Line of action for the 10 lb weight is $x = 1$

Let the point of intersection for the lines of action for $T_1$ and $T_2$ be the pivot point $P$.

Since $T_1$ and $T_2$ act through point $P$, they exert no torque relative to point $P$.

Relative to point $P$, the 10 lb weight exerts a clockwise torque, $\tau = \dfrac{10}{3} \text{ ft-lbs}$.

To counteract that torque, a weight must be hung that exerts a torque of $\dfrac{10}{3} \text{ ft-lbs}$ in the counterclockwise direction. The least weight would be a 5 lb weight hung on the left edge of the picture frame $\dfrac{2}{3} \text{ ft}$ to the left of point $P$.
I’m really thankful to you for answering me but I have few misconceptions:-

1. Why do we need to find the intersection of $$\displaystyle T_1$$ and $$\displaystyle T_2$$ ? I mean is it done just for solving the problem or there was a real need for that.

2. Was it a guess that a $$\displaystyle \frac {10} {3}$$ weight would be $$\displaystyle \frac{2}{3} ft$$ to the left because we could also say that that the balancing weight of $$\displaystyle \frac{10} {3} lb$$ is 1 ft to the left of the pivot point P. Please correct me if I’m wrong.

Using a Cartesian grid (see attached diagram), let the origin be the point where $T_1$ meets the upper left corner of the picture frame.

Line of action for $T_1$ is $y=-x$

Line of action for $T_2$ is $y=\dfrac{x}{2}-1$

Point of intersection for these two lines of action is $\left(\dfrac{2}{3},-\dfrac{2}{3}\right)$

Line of action for the 10 lb weight is $x = 1$

Let the point of intersection for the lines of action for $T_1$ and $T_2$ be the pivot point $P$.

Since $T_1$ and $T_2$ act through point $P$, they exert no torque relative to point $P$.

Relative to point $P$, the 10 lb weight exerts a clockwise torque, $\tau = \dfrac{10}{3} \text{ ft-lbs}$.

To counteract that torque, a weight must be hung that exerts a torque of $\dfrac{10}{3} \text{ ft-lbs}$ in the counterclockwise direction. The least weight would be a 5 lb weight hung on the left edge of the picture frame $\dfrac{2}{3} \text{ ft}$ to the left of point $P$.
Thank you for answering my question. I have few doubts regarding your solution

1. Why we have taken the intersection of line of action of tension forces? I mean is it done just for solving this particular problem or is there any real need for it? We will get a net torque of $$\displaystyle -10/3 lb$$ if we assume $$\displaystyle \Sigma F_y =0$$ and $$\displaystyle \Sigma F_x = 0$$ and calculating the net torque by taking the pivot point to be the center of the picture.

2. Why we have taken $$\displaystyle 2/3 ft.$$ left to the pivot point? I mean we could take $$\displaystyle 1 ft$$ to the left of pivot point and a weight of $$\displaystyle 10/3 lb$$ without any restrictions.

I hope my doubts will be cleared.

I’m thankful to you for your answer. But please tell me why you have taken the intersection of two strings, I mean is there some reason for it and it is done just for solving this particular problem?

#### Woody

Cervesa mentions "lines of action".

These are "imagined" lines which follow the lines of tension,
from the nail, along the strings, through the corner of the painting
and onward until the two lines meet.

Reducing the problem to its minimum features,
allows the solution to be seen without being distracted by extraneous detail.

1 person

What was the purpose of extending those lines of action? I mean why can’t we add those lines of force vectorially?

#### Woody

oops

Shows the dangers of skim reading (and assuming that you know what you are talking about)

You are of course correct
Cervesa is actually doing the vector calculation as you suggest.

This computer (apart from having no sound card) does not display the maths equations properly:
(I just get an "undefined control sequence" error)

From a quick glance at what I could see of the post, and a squint at the attached picture,
I jumped to a conclusion.

1 person