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 ).
I request you all to please help me over here. I know all the laws that applies in statics but then also I can't solve them, please help me with this too.
Thank you any help will be much appreciated.
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 ).
I request you all to please help me over here. I know all the laws that applies in statics but then also I can't solve them, please help me with this too.
Thank you any help will be much appreciated.
Attachments

797.1 KB Views: 1

63.9 KB Views: 1
Last edited: