Torque!!!

[peacefulheart]

A sign hangs by a rope attached at 3D" to the middle of its
upper edge. It rests against a frictionless wall. If the weight
of the sign were doubled. what would happen to the tension
in the string? (Note: sin 30° =0.5; cos 30° =0.87).

A. It would remain the same.
B. It would increase by a factor of 1.5.
C. It would increase by a factor of 2.
D. It would increase by a factor of 4.

1. I have no problem with answer C

2. I am confused about the answer explanation . It says " Since the wall is frictionless, there is no torque in this problem."

3. I think the net torque is zero since the torque exerted by vertical component of the string tension is cancelled by the torque exerted by the weight of sign.


thanks a lot

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[Captain Sisko]

A sign hangs by a rope attached at 3D" to the middle of its
upper edge. It rests against a frictionless wall. If the weight
of the sign were doubled. what would happen to the tension
in the string? (Note: sin 30° =0.5; cos 30° =0.87).

A. It would remain the same.
B. It would increase by a factor of 1.5.
C. It would increase by a factor of 2.
D. It would increase by a factor of 4.

1. I have no problem with answer C

2. I am confused about the answer explanation . It says " Since the wall is frictionless, there is no torque in this problem."

3. I think the net torque is zero since the torque exerted by vertical component of the string tension is cancelled by the torque exerted by the weight of sign.


thanks a lot

First, you need to post the picture.

If the wall had friction it would apply a vertical force on the sign, making the tension in the rope less than what you calculated. If it did have friction, the only way you could solve the problem is by using a net moment equation which would include torque.

 

[contraband]

I think the sign looks something like figure 1 in the attached ppt.

Now, we know the sign is not accelerating either translationally or rotationally - it's at static equilibrium.

That means the net torque and net force must be zero, and that either force or torque can be used to solve. I find that this problem is easiest to solve using force (if you get stuck trying to solve something one way with a static equilibrium problem, try the other approach!):

For any problem involving force, I like to make two free body diagrams - one for the net force, and one for the component forces, as seen in figure 2.

Fnet = 0 , so it's just a dot.

The first component forces I think of are tension and gravity, so I draw them in for figure 2. In figure 3, I break T into x and y components since that makes it easier to make components add to 0, and it shows me that I need another force in the x direction to balance out Tx - that's the normal force from the wall.

What is the problem changing? mg - the weight!
mg = Ty
Ty = Tsin(30) = 0.5T

Thus, mg = 0.5T

mg is directly proportional to T, meaning that if I double mg, T is doubled as well.

The explanation's a little misleading - there is definitely torque, but there's no NET torque. The fact that the wall's frictionless just takes away one type of torque that could have been applied.

 

[peacefulheart]

thanks a lot for the detailed explanation.

 

[peacefulheart]

I think the sign looks something like figure 1 in the attached ppt.

Now, we know the sign is not accelerating either translationally or rotationally - it's at static equilibrium.

That means the net torque and net force must be zero, and that either force or torque can be used to solve. I find that this problem is easiest to solve using force (if you get stuck trying to solve something one way with a static equilibrium problem, try the other approach!):

For any problem involving force, I like to make two free body diagrams - one for the net force, and one for the component forces, as seen in figure 2.

Fnet = 0 , so it's just a dot.

The first component forces I think of are tension and gravity, so I draw them in for figure 2. In figure 3, I break T into x and y components since that makes it easier to make components add to 0, and it shows me that I need another force in the x direction to balance out Tx - that's the normal force from the wall.

What is the problem changing? mg - the weight!
mg = Ty
Ty = Tsin(30) = 0.5T

Thus, mg = 0.5T

mg is directly proportional to T, meaning that if I double mg, T is doubled as well.

The explanation's a little misleading - there is definitely torque, but there's no NET torque. The fact that the wall's frictionless just takes away one type of torque that could have been applied.

Let suppose the wall is not frictionless. Then the direction of friction is vertical. If we choose the point where the sign and the wall attach as pivot point, then the torque is till going to be zero since there is no r.


thanks a lot

 

[contraband]

Exactly, so in either case the torque would sum to zero - the wall being frictionless just takes away one kind of component torque.

 

[peacefulheart]

Exactly, so in either case the torque would sum to zero - the wall being frictionless just takes away one kind of component torque.

thanks.

That is why I get confused about the answer explanation "Since the wall is frictionless, there is no torque in this problem".