What's wrong with my approach to drawing this model?
$begingroup$
An object of mass 1.8kg is attached to the ends of two light elastic strings having the same modulus of elasticity. One of the strings has natural length 0.8m and the other has a natural length of 1.1m. The longer string is attached at A and the shorter string is attached at B on the same horizontal level. The object hangs 0.85m below O, a point on the same level as A and B, 1.4m from A and 0.8m from B. Find the modulus of elasticity of the strings.
I'm having a bit of trouble with modelling the question with a diagram. I know what the correct answer is, and the model needed to help solve it (1) (bear in mind I've omitted some info to draw them faster), but I thought (2) will also be valid, just using the extension of the string at B as 0.05 instead of using Pythagoras' theorem.
Why not 2?
proof-verification physics classical-mechanics
edited Dec 26 '18 at 20:49
Andrei
13.1k21230
asked Dec 26 '18 at 19:48
Gab N.Gab N.
483
$endgroup$
add a comment |
$begingroup$
An object of mass 1.8kg is attached to the ends of two light elastic strings having the same modulus of elasticity. One of the strings has natural length 0.8m and the other has a natural length of 1.1m. The longer string is attached at A and the shorter string is attached at B on the same horizontal level. The object hangs 0.85m below O, a point on the same level as A and B, 1.4m from A and 0.8m from B. Find the modulus of elasticity of the strings.
I'm having a bit of trouble with modelling the question with a diagram. I know what the correct answer is, and the model needed to help solve it (1) (bear in mind I've omitted some info to draw them faster), but I thought (2) will also be valid, just using the extension of the string at B as 0.05 instead of using Pythagoras' theorem.
Why not 2?
proof-verification physics classical-mechanics
edited Dec 26 '18 at 20:49
Andrei
13.1k21230
asked Dec 26 '18 at 19:48
Gab N.Gab N.
483
$endgroup$
add a comment |
$begingroup$
An object of mass 1.8kg is attached to the ends of two light elastic strings having the same modulus of elasticity. One of the strings has natural length 0.8m and the other has a natural length of 1.1m. The longer string is attached at A and the shorter string is attached at B on the same horizontal level. The object hangs 0.85m below O, a point on the same level as A and B, 1.4m from A and 0.8m from B. Find the modulus of elasticity of the strings.
I'm having a bit of trouble with modelling the question with a diagram. I know what the correct answer is, and the model needed to help solve it (1) (bear in mind I've omitted some info to draw them faster), but I thought (2) will also be valid, just using the extension of the string at B as 0.05 instead of using Pythagoras' theorem.
Why not 2?
proof-verification physics classical-mechanics
edited Dec 26 '18 at 20:49
Andrei
13.1k21230
asked Dec 26 '18 at 19:48
Gab N.Gab N.
483
$endgroup$
An object of mass 1.8kg is attached to the ends of two light elastic strings having the same modulus of elasticity. One of the strings has natural length 0.8m and the other has a natural length of 1.1m. The longer string is attached at A and the shorter string is attached at B on the same horizontal level. The object hangs 0.85m below O, a point on the same level as A and B, 1.4m from A and 0.8m from B. Find the modulus of elasticity of the strings.
I'm having a bit of trouble with modelling the question with a diagram. I know what the correct answer is, and the model needed to help solve it (1) (bear in mind I've omitted some info to draw them faster), but I thought (2) will also be valid, just using the extension of the string at B as 0.05 instead of using Pythagoras' theorem.
Why not 2?
proof-verification physics classical-mechanics
proof-verification physics classical-mechanics
edited Dec 26 '18 at 20:49
Andrei
13.1k21230
asked Dec 26 '18 at 19:48
Gab N.Gab N.
483
edited Dec 26 '18 at 20:49
Andrei
13.1k21230
asked Dec 26 '18 at 19:48
Gab N.Gab N.
483
edited Dec 26 '18 at 20:49
Andrei
13.1k21230
edited Dec 26 '18 at 20:49
Andrei
13.1k21230
edited Dec 26 '18 at 20:49
Andrei
13.1k21230
13.1k21230
asked Dec 26 '18 at 19:48
Gab N.Gab N.
483
asked Dec 26 '18 at 19:48
Gab N.Gab N.
483
asked Dec 26 '18 at 19:48
Gab N.Gab N.
483
483
add a comment |
add a comment |
1 Answer
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(2) is not valid. The reason is that (2) is not an equilibrium case. The string to $A$ has a horizontal component of the elastic force. In order to be in equilibrium, the elastic force of the string to $B$ has to have the horizontal component of the same magnitude, in opposite direction
answered Dec 26 '18 at 19:52
AndreiAndrei
13.1k21230
$endgroup$
$begingroup$
Oh, okay. But why does the system have to be in equilibrium?
$endgroup$
– Gab N.
Dec 26 '18 at 20:45
$begingroup$
If it's not in the equilibrium, it means that the sum of the forces is not zero, so the object will accelerate. "Hangs" implies not moving, so no acceleration.
$endgroup$
– Andrei
Dec 26 '18 at 20:49
$begingroup$
Note that the real answer is somewhere between the symmetric case (1) and the case(2). So the angle the two strings make with the vertical are not the same.
$endgroup$
– Andrei
Dec 26 '18 at 20:51
$begingroup$
Ahh, right, I get it now! Thank you!
$endgroup$
– Gab N.
Dec 26 '18 at 21:11
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
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$begingroup$
(2) is not valid. The reason is that (2) is not an equilibrium case. The string to $A$ has a horizontal component of the elastic force. In order to be in equilibrium, the elastic force of the string to $B$ has to have the horizontal component of the same magnitude, in opposite direction
answered Dec 26 '18 at 19:52
AndreiAndrei
13.1k21230
$endgroup$
$begingroup$
Oh, okay. But why does the system have to be in equilibrium?
$endgroup$
– Gab N.
Dec 26 '18 at 20:45
$begingroup$
If it's not in the equilibrium, it means that the sum of the forces is not zero, so the object will accelerate. "Hangs" implies not moving, so no acceleration.
$endgroup$
– Andrei
Dec 26 '18 at 20:49
$begingroup$
Note that the real answer is somewhere between the symmetric case (1) and the case(2). So the angle the two strings make with the vertical are not the same.
$endgroup$
– Andrei
Dec 26 '18 at 20:51
$begingroup$
Ahh, right, I get it now! Thank you!
$endgroup$
– Gab N.
Dec 26 '18 at 21:11
add a comment |
$begingroup$
(2) is not valid. The reason is that (2) is not an equilibrium case. The string to $A$ has a horizontal component of the elastic force. In order to be in equilibrium, the elastic force of the string to $B$ has to have the horizontal component of the same magnitude, in opposite direction
answered Dec 26 '18 at 19:52
AndreiAndrei
13.1k21230
$endgroup$
$begingroup$
Oh, okay. But why does the system have to be in equilibrium?
$endgroup$
– Gab N.
Dec 26 '18 at 20:45
$begingroup$
If it's not in the equilibrium, it means that the sum of the forces is not zero, so the object will accelerate. "Hangs" implies not moving, so no acceleration.
$endgroup$
– Andrei
Dec 26 '18 at 20:49
$begingroup$
Note that the real answer is somewhere between the symmetric case (1) and the case(2). So the angle the two strings make with the vertical are not the same.
$endgroup$
– Andrei
Dec 26 '18 at 20:51
$begingroup$
Ahh, right, I get it now! Thank you!
$endgroup$
– Gab N.
Dec 26 '18 at 21:11
add a comment |
$begingroup$
(2) is not valid. The reason is that (2) is not an equilibrium case. The string to $A$ has a horizontal component of the elastic force. In order to be in equilibrium, the elastic force of the string to $B$ has to have the horizontal component of the same magnitude, in opposite direction
answered Dec 26 '18 at 19:52
AndreiAndrei
13.1k21230
$endgroup$
(2) is not valid. The reason is that (2) is not an equilibrium case. The string to $A$ has a horizontal component of the elastic force. In order to be in equilibrium, the elastic force of the string to $B$ has to have the horizontal component of the same magnitude, in opposite direction
answered Dec 26 '18 at 19:52
AndreiAndrei
13.1k21230
answered Dec 26 '18 at 19:52
AndreiAndrei
13.1k21230
answered Dec 26 '18 at 19:52
AndreiAndrei
13.1k21230
answered Dec 26 '18 at 19:52
AndreiAndrei
13.1k21230
13.1k21230
$begingroup$
Oh, okay. But why does the system have to be in equilibrium?
$endgroup$
– Gab N.
Dec 26 '18 at 20:45
$begingroup$
If it's not in the equilibrium, it means that the sum of the forces is not zero, so the object will accelerate. "Hangs" implies not moving, so no acceleration.
$endgroup$
– Andrei
Dec 26 '18 at 20:49
$begingroup$
Note that the real answer is somewhere between the symmetric case (1) and the case(2). So the angle the two strings make with the vertical are not the same.
$endgroup$
– Andrei
Dec 26 '18 at 20:51
$begingroup$
Ahh, right, I get it now! Thank you!
$endgroup$
– Gab N.
Dec 26 '18 at 21:11
add a comment |
$begingroup$
Oh, okay. But why does the system have to be in equilibrium?
$endgroup$
– Gab N.
Dec 26 '18 at 20:45
$begingroup$
If it's not in the equilibrium, it means that the sum of the forces is not zero, so the object will accelerate. "Hangs" implies not moving, so no acceleration.
$endgroup$
– Andrei
Dec 26 '18 at 20:49
$begingroup$
Note that the real answer is somewhere between the symmetric case (1) and the case(2). So the angle the two strings make with the vertical are not the same.
$endgroup$
– Andrei
Dec 26 '18 at 20:51
$begingroup$
Ahh, right, I get it now! Thank you!
$endgroup$
– Gab N.
Dec 26 '18 at 21:11
$begingroup$
Oh, okay. But why does the system have to be in equilibrium?
$endgroup$
– Gab N.
Dec 26 '18 at 20:45
$begingroup$
Oh, okay. But why does the system have to be in equilibrium?
$endgroup$
– Gab N.
Dec 26 '18 at 20:45
$begingroup$
If it's not in the equilibrium, it means that the sum of the forces is not zero, so the object will accelerate. "Hangs" implies not moving, so no acceleration.
$endgroup$
– Andrei
Dec 26 '18 at 20:49
$begingroup$
If it's not in the equilibrium, it means that the sum of the forces is not zero, so the object will accelerate. "Hangs" implies not moving, so no acceleration.
$endgroup$
– Andrei
Dec 26 '18 at 20:49
$begingroup$
Note that the real answer is somewhere between the symmetric case (1) and the case(2). So the angle the two strings make with the vertical are not the same.
$endgroup$
– Andrei
Dec 26 '18 at 20:51
$begingroup$
Note that the real answer is somewhere between the symmetric case (1) and the case(2). So the angle the two strings make with the vertical are not the same.
$endgroup$
– Andrei
Dec 26 '18 at 20:51
$begingroup$
Ahh, right, I get it now! Thank you!
$endgroup$
– Gab N.
Dec 26 '18 at 21:11
$begingroup$
Ahh, right, I get it now! Thank you!
$endgroup$
– Gab N.
Dec 26 '18 at 21:11
add a comment |
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