What Did I Miss in This Torque Equilibrium Problem?

[Adam Rabe]

Homework Statement

Homework Equations

T = fd
f = mg

The Attempt at a Solution

I appear to be suffering from autism.
Heres my working out...

F(bar) = 200
Two support forces of equal length, each 0.5 m from fulcrum (given). Therefore torque magnitude is the same. 100 N on both sides = counters 200 downwards. My answer is therefore B.

Answer sheet says that A.) is the right answer.
Can someone help me out, what did i miss? Thanks in advance for the help

 

[kuruman]

Perhaps you misunderstood the problem. You have two supports at 1 m and 2 m from the 0 m end. You are required to find what force (not torque) must be applied at the 0 m end to lift that end off the supports. What condition must be true for this to happen?

 

[SammyS]

Homework Statement

View attachment 204211

Homework Equations

T = fd
f = mg

The Attempt at a Solution

I appear to be suffering from autism.
Heres my working out...

F(bar) = 200
Two support forces of equal length, each 0.5 m from fulcrum (given). Therefore torque magnitude is the same. 100 N on both sides = counters 200 downwards. My answer is therefore B.

Answer sheet says that A.) is the right answer.
Can someone help me out, what did i miss? Thanks in advance for the help

Here is the figure from the "Problem Statement" section of your post.

This may aid in the willingness of people to help with this problem.

 

[scottdave]

Take each section and calculate the torque of that section. Say I am going to try pushing up on the left end. The support closest to me will provide upward force (and therefore torque) until I am starting to lift (when it is no longer in contact), so the support closest to me can be ignored to calculate this problem.

There is no middle fulcrum, like a balanced seesaw. The far support is the pivot point. So I am 2 meters away from the pivot point. This is kind of like a wheelbarrow. Initially it is supported by the wheel and the back legs, but when you pick it up, you only have a pivot on the wheel. Does this help? I ran through the calculations and concur with the answer key.

 

[kuruman]

I ran through the calculations and concur with the answer key.

Me too. However, to see what's going on and why, you still need to answer my question, "what must be true for the rod to lift off?" Specifically, what is the force exerted by the support closer to where the upward lifting force is applied?

 

[scottdave]

As you start to apply an upward force, the support near you will exert less force, so that your force plus the two supports add up to 200 Newtons. At the moment that you lift it off, the support (nearest you) is providing zero. So you must provide greater than or equal to the amount which will support it as if there was not a "nearest support".

The problem statement says nothing about a fulcrum in the center. In fact, it tells you that there are only two places that support the beam (1 meter from each end). You cannot make any solution which uses a "middle fulcrum".

 

[Adam Rabe]

Me too. However, to see what's going on and why, you still need to answer my question, "what must be true for the rod to lift off?" Specifically, what is the force exerted by the support closer to where the upward lifting force is applied?

Hello sorry for the late reply.

Ok i see now that there are two support forces. For the rod to lift off net force upwards must be greater than downwards. The two support forces must still be providing 100 each (is this right?)
If i lift one end, the other end would still be providing a support force of 100.
The torque force applied to the 0m end should be equal to the torque force of 100 * 2 m (2m = distance from other end)
and so...
100*2=200 = F(3m)
F = 66.66
Is this right? i realize I am 16N off

 

[haruspex]

If i lift one end, the other end would still be providing a support force of 100.

You cannot assume that. Consider it unknown.

The torque force applied to the 0m end should be equal to the torque force of 100 * 2 m (2m = distance from other end)

I can't follow what you are doing there. Be more precise about which forces you are considering and what point you are taking moments about.

In general, in 2D statics, you have three equations available:
ΣF=0 in each of two directions (here there are no horizontal forces so you have only one interesting equation)
Στ=0, where the torques are all taken about some specified axis.
What axis do you choose? In principle, the choice does not matter as long as you are consistent, but if any of the forces are uninteresting (not given and not asked for) you can avoid involving them by careful choice of axis.

 

[Adam Rabe]

You cannot assume that. Consider it unknown.

I can't follow what you are doing there. Be more precise about which forces you are considering and what point you are taking moments about.

In general, in 2D statics, you have three equations available:
ΣF=0 in each of two directions (here there are no horizontal forces so you have only one interesting equation)
Στ=0, where the torques are all taken about some specified axis.
What axis do you choose? In principle, the choice does not matter as long as you are consistent, but if any of the forces are uninteresting (not given and not asked for) you can avoid involving them by careful choice of axis.

Yes it does. Try balancing a pen on two places and lifting one end up, the other end is still supported.

 

[scottdave]

Yes it does. Try balancing a pen on two places and lifting one end up, the other end is still supported.

He meant that you cannot assume that the other support is still providing the same 100 N force. In your pen example, if the pen 'weighs' 2 ounces, and at first each finger provides 1 ounce force. Then you start providing more upward force with one finger, the total upward force is still the same 2 ounces. So if one finger is pushing up with 1.5 ounces, then the other is 0.5 ounces (unless you are accelerating the pen).

 

[scottdave]

Is this for a class? If so, I think now would be the time to go ask help from a TA or your Prof, or a helpdesk or fellow student.
What I am saying, is that it seems we have reached the limit of being able to convey the concepts of this problem, effectively (on this platform). When you are talking with a person "live" you can get immediate feedback.

 

[Adam Rabe]

You cannot assume that. Consider it unknown.

I can't follow what you are doing there. Be more precise about which forces you are considering and what point you are taking moments about.

In general, in 2D statics, you have three equations available:
ΣF=0 in each of two directions (here there are no horizontal forces so you have only one interesting equation)
Στ=0, where the torques are all taken about some specified axis.
What axis do you choose? In principle, the choice does not matter as long as you are consistent, but if any of the forces are uninteresting (not given and not asked for) you can avoid involving them by careful choice of axis.

Hey is this easier to follow?
Am i on the right track

 

[haruspex]

Then you start providing more upward force with one finger

Actually, that is not the point, unless you are providing non-negligible acceleration (and in this question it should be considered negligible).
The point is that one upward force is being applied at a different distance from the mass centre than previously. That changes the distribution of the support forces, even with no movement.

 

[Adam Rabe]

Actually, that is not the point, unless you are providing non-negligible acceleration (and in this question it should be considered negligible).
The point is that one upward force is being applied at a different distance from the mass centre than previously. That changes the distribution of the support forces, even with no movement.

I understand.
So how would you go about calculating these shifts in distances and size of support forces? I wasn't really taught how to do so, my lecturers kinda just threw me the torque formula.
Is there some algebraic method i need to do?

 

[haruspex]

Am i on the right track

No.
You wrote "F=3m=torque up = 200"
I assume you meant "F x 3m=torque up = 200", but the 200 is a force (N) not a torque (Nm).
And you have still not answered my question, what point are you taking as axis for torque?
From the "F x 3m", it looks like you are taking the right hand end of the rod as the axis. If so:
- what is the torque of the 200N about that axis?
- what is the torque from the other support about that axis?

 

[Adam Rabe]

I don't quite understand what you mean by about that axis.
Do you mean like making the weight force the centre of the two support forces?
So... have the weight force in between support force A and B
Left as fulcrum: Torque = 0 = (-200 * 0.5) + (Fb * 1) <-- support force by other side of axis (B) = 100
Right as fulcrum: Torque = 0 = (-200 * 0.5) + (Fa * 1) <-- support force by other side of axis (A) = 100

So 100?

 

[haruspex]

I don't quite understand what you mean by about that axis.

It means nothing to ask what torque a force exerts unless you specify the axis. You have to choose one, and state what point you have chosen.
In a simple set-up like this, with all forces vertical and all points of application of those forces in a straight horizontal line, it is obvious to choose an axis somewhere on the rod.
Having chosen the axis, the torque a force exerts about the axis is proportional to its distance from the axis.
You have to be careful to distinguish clockwise from anticlockwise. An upward force F acting distance x to the left of the axis exerts clockwise torque Fx. The same force acting distance x to the right of the axis exerts anticlockwise torque Fx.
The usual convention is positive for upwards forces, positive for displacements to the right, and positive for anticlockwise torques.
E.g. if you take the right hand end as axis then the weight exerts a force -200N at displacement -1.5m so exerts a torque +300Nm.

Left as fulcrum: Torque = 0 = (-200 * 0.5) + (Fb * 1) <-- support force by other side of axis (B) = 100

If by "left as fulcrum" you mean the point 1m from left hand end as axis, then yes.
But that is for the initial situation, with the rod resting on the symmetrically placed supports. What happens when one of those is replaced by a support at the end of the rod?

 

[Adam Rabe]

It means nothing to ask what torque a force exerts unless you specify the axis. You have to choose one, and state what point you have chosen.
In a simple set-up like this, with all forces vertical and all points of application of those forces in a straight horizontal line, it is obvious to choose an axis somewhere on the rod.
Having chosen the axis, the torque a force exerts about the axis is proportional to its distance from the axis.
You have to be careful to distinguish clockwise from anticlockwise. An upward force F acting distance x to the left of the axis exerts clockwise torque Fx. The same force acting distance x to the right of the axis exerts anticlockwise torque Fx.
The usual convention is positive for upwards forces, positive for displacements to the right, and positive for anticlockwise torques.
E.g. if you take the right hand end as axis then the weight exerts a force -200N at displacement -1.5m so exerts a torque +300Nm.If by "left as fulcrum" you mean the point 1m from left hand end as axis, then yes.
But that is for the initial situation, with the rod resting on the symmetrically placed supports. What happens when one of those is replaced by a support at the end of the rod?

Then the distance would of the left side would be longer... just like on my drawing
so point A from weight force would be 1m and point B from weight force would still be 0.5m

Left as fulcrum: Torque = 0 = (-200 * 1m) + (Fb * 1.5) => Fb = 200/1.5
Right as fulcrum: Torque = 0 = (-200 *0.5) + (Fa * 1.5) => Fb = 100/1.5 = 66.66

So was i right?? but I am still 16 N off

 

[haruspex]

Left as fulcrum

Despite all my hints and prods, I do not seem to be able to get you to state clearly what you mean.
What point do you mean here? The left hand end of the rod or the 1m from left position of the original support?
Similarly "Right as fulcrum"

 

[Adam Rabe]

Despite all my hints and prods, I do not seem to be able to get you to state clearly what you mean.
What point do you mean here? The left hand end of the rod or the 1m from left position of the original support?
Similarly "Right as fulcrum"

Say two support forces from left to right are A and B.
I meant that when 'Left is Fulcrum'. left support force (A) is the pivot point of system. Right is fulcrum, right support force (B) is the pivot point of system

 

[haruspex]

Say two support forces from left to right are A and B.
I meant that when 'Left is Fulcrum'. left support force (A) is the pivot point of system. Right is fulcrum, right support force (B) is the pivot point of system

Ok, so now we have A at 0m from left, weight at 1.5m from left, support B at 2m from left, correct?

Taking point A as axis you write:

0 = (-200 * 1m) + (Fb * 1.5)

How do you get the 1m and 1.5m?

Taking point B as axis you write:

0 = (-200 *0.5) + (Fa * 1.5)

How do you get the 1.5m?

 

[Adam Rabe]

Ok, so now we have A at 0m from left, weight at 1.5m from left, support B at 2m from left, correct?

Taking point A as axis you write:

How do you get the 1m and 1.5m?

Taking point B as axis you write:

How do you get the 1.5m?

Hi sorry this is taking up a lot of your time. I really appreciate your help
Does this image make sense? its essentially just like the previous drawing i did

 

[haruspex]

Does this image make sense?

The second diagram is wrong. I think you have confused yourself by not drawing the whole bar in each picture.
The bar is 3m long. The weight is in the middle. In the situation to be analysed, point A is at the very left hand end of the bar. How far does that make it from the weight?

 

[kuruman]

Perhaps a couple of good pictures will help. The ones below are drawn to scale.

Picture on the left
It shows the forces when the rod is supported and is at equilibrium as given by the problem. Note that the forces exerted by the supports are obviously equal to 100 N each. This means that if you were to apply an upward force of 100 N at the left support or 1.0 m away from the left end, the rod will be just about on the verge of being lifted off.

Picture on the right
It shows the forces as if the rod is supported and is at equilibrium with the right support untouched but the left support moved to the end. Clearly the forces needed to keep the rod at equilibrium cannot be 100 N each. The force F that is needed to keep equilibrium at 0 m is the force that you are asked to find.