Proving tension in rope of equilibrium system

[toforfiltum]

Homework Statement

Homework Equations

principle of moments

The Attempt at a Solution

Okay so when resolving the forces horizontally and vertically, I get R_A= T cosθ for horizontal component and R_B= W + T sinθ for vertical component. To get my T, I chose to take moments about B, though I'm not sure if it's the fastest way. So 2 R_A = ¾W + T sinθ( 1.5 - OC cosθ).

I get the ¾ value by using similar triangles. Now, the problem I have is eliminating OC from my equation, because OC is not perpendicular to AB, which leaves me with many unknowns that I don't know how to eliminate.
So, can someone help me find a way to eliminate OC?

 

[SammyS]

Homework Statement

View attachment 86514

Homework Equations

principle of moments

The Attempt at a Solution

Okay so when resolving the forces horizontally and vertically, I get R_A= T cosθ for horizontal component and R_B= W + T sinθ for vertical component. To get my T, I chose to take moments about B, though I'm not sure if it's the fastest way. So 2 R_A = ¾W + T sinθ( 1.5 - OC cosθ).

I get the ¾ value by using similar triangles. Now, the problem I have is eliminating OC from my equation, because OC is not perpendicular to AB, which leaves me with many unknowns that I don't know how to eliminate.
So, can someone help me find a way to eliminate OC?

You are missing a contribution to the torque. The horizontal component of T also exerts torque about point B.

Also: You will need to do some trig stuff to eliminate length OC .

 

[toforfiltum]

You are missing a contribution to the torque. The horizontal component of T also exerts torque about point B.

Also: You will need to do some trig stuff to eliminate length OC .

You are missing a contribution to the torque. The horizontal component of T also exerts torque about point B.

Also: You will need to do some trig stuff to eliminate length OC .

Okay, so I rewrite my equation as 2 R_A = ¾W + Tsinθ (1.5-OC cosθ) + Tcosθ (2-OC sinθ)
Since R_A= T cosθ, I substituted this value into the equation above.
Then. after some algebraic manipulation, I still did not manage to eliminate my OC, I get T= 3W / ( 8 OC cosθsinθ- 6 sinθ), which looks a bit like the answer, but isn't.
Must I form another equation to eliminate OC?

 

[andrewkirk]

You are given all angles in the diagram, plus $W$, so you have three unknowns: the forces of the wall on the ladder (horizontal), the floor on the ladder (vertical) and the rope on the ladder (at the angle $\theta$).

So you need three equations to solve it. The torque gives you one equation that is required in order for the ladder not to rotate. You have written that (I have not checked it). You can get the other two equations by requiring the vertical and horizontal components of force on the ladder to be zero, so that it does not move.

 

[toforfiltum]

You are given all angles in the diagram, plus $W$, so you have three unknowns: the forces of the wall on the ladder (horizontal), the floor on the ladder (vertical) and the rope on the ladder (at the angle $\theta$).

So you need three equations to solve it. The torque gives you one equation that is required in order for the ladder not to rotate. You have written that (I have not checked it). You can get the other two equations by requiring the vertical and horizontal components of force on the ladder to be zero, so that it does not move.

I have the three equations already, but I still can't manage to eliminate OC from my equation. That's my main problem. Now I'm not sure if it's necessary to include the OC in my moment equation, since I cannot eliminate it. But I can't find a better way, since its hard to find the perpendicular distance of tension from B, with OC not perpendicular to AB

 

[andrewkirk]

I notice a couple of things about your equation:

1. I think you should have $\frac{3}{5}W$ not $\frac{3}{4}W$ for the torque about B provided by the ladder's weight.
2. I don't think there's any need to put OC in your equation. You have resolved the rope's tension into horizontal and vertical components. Now resolve each of those into components parallel and perpendicular to the ladder, using the 3:4:5 triangle geometry. Add the two perpendicular ones, multiply by BC and you have the contribution to torque from the rope. To get BC, drop a perp from C to OB to divide OCB into two triangles. The geom of one of those is given by $\theta$, the other is 3:4:5, and you know OB=1.5m, so you can express BC in terms of $\theta$.

 

[SammyS]

I notice a couple of things about your equation:

1. I think you should have $\frac{3}{5}W$ not $\frac{3}{4}W$ for the torque about B provided by the ladder's weight.

(3/4)W is correct.

 

[andrewkirk]

(3/4)W is correct.

Ah yes. The force that generates the torque is $\frac{3}{5}W$. I neglected to multiply by the distance of $\frac{5}{4}m$ from ladder's CoM to $B$, which makes the torque $\frac{3}{4}W$.

 

[toforfiltum]

Ah yes. The force that generates the torque is $\frac{3}{5}W$. I neglected to multiply by the distance of $\frac{5}{4}m$ from ladder's CoM to $B$, which makes the torque $\frac{3}{4}W$.

I still can't manage to express BC in terms of θ only. Here are my steps:

tanθ = x/ (1.5-y), where x and y are two unknowns.
tan α = x/y , where α is part of the 3:4:5 triangle.

Since both x are the same, tanα = tanθ (1.5-y)/ y
which gives 4/3 = tanθ (1.5-y)/ y

cosα=y/ BC
BC = ¾ tanθ (1.5-y) x 5/3
=5/4 tanθ (1.5-y)

I've tried for quite some time, and I can't eliminate the y from my equation. This question has exasperated me for hours.

 

[toforfiltum]

Ah yes. The force that generates the torque is $\frac{3}{5}W$. I neglected to multiply by the distance of $\frac{5}{4}m$ from ladder's CoM to $B$, which makes the torque $\frac{3}{4}W$.

I still can't manage to express BC in terms of θ only. Here are my steps:

tanθ = x/ (1.5-y), where x and y are two unknowns.
tan α = x/y , where α is part of the 3:4:5 triangle.

Since both x are the same, tanα = tanθ (1.5-y)/ y
which gives 4/3 = tanθ (1.5-y)/ y

cosα=y/ BC
BC = ¾ tanθ (1.5-y) x 5/3
=5/4 tanθ (1.5-y)

I've tried for quite some time, and I can't eliminate the y from my equation. This question has exasperated me for hours.

 

[toforfiltum]

Ah yes. The force that generates the torque is $\frac{3}{5}W$. I neglected to multiply by the distance of $\frac{5}{4}m$ from ladder's CoM to $B$, which makes the torque $\frac{3}{4}W$.

I still can't manage to express BC in terms of θ only. Here are my steps:

tanθ = x/ (1.5-y), where x and y are two unknowns.
tan α = x/y , where α is part of the 3:4:5 triangle.

Since both x are the same, tanα = tanθ (1.5-y)/ y
which gives 4/3 = tanθ (1.5-y)/ y

cosα=y/ BC
BC = ¾ tanθ (1.5-y) x 5/3
=5/4 tanθ (1.5-y)

I've tried for quite some time, and I can't eliminate the y from my equation. This question has exasperated me for hours.

 

[toforfiltum]

Sorry about the multiple posts...internet was laggy, but I can't delete them.

 

[Qwertywerty]

Why don't you try balancing torque from point O ?

Now you wouldn't have to worry about OC or anything else .

Hope this helps .

 

[toforfiltum]

Why don't you try balancing torque from point O ?

Now you wouldn't have to worry about OC or anything else .

Hope this helps .

I don't see how balancing torque from point O eliminates the need to use OC. How about the perpendicular distance from horizontal component of T to O, and also for the vertical component?

 

[Qwertywerty]

I don't see how balancing torque from point O eliminates the need to use OC. How about the perpendicular distance from horizontal component of T to O, and also for the vertical component?

Tension will not produce a torque about point O . What is the line of action of tension ?

 

[toforfiltum]

Tension will not produce a torque about point O . What is the line of action of tension ?

So if O lies in the line of action of tension, then there would be no moment. Hmm, I see that I'm wrong. Thanks for pointing it out.

 

[Qwertywerty]

So if O lies in the line of action of tension, then there would be no moment. Hmm, I see that I'm wrong. Thanks for pointing it out.

So , can you now solve this question ?

 

[toforfiltum]

So , can you now solve this question ?

Wow, just wow. I'm speechless. After trying for so long and wasting so much time, you made it look so easy, and I feel relieved and rather dumb at the same time. Many many thanks to you, this is one of the few questions that have managed to stump me for a long time.

By the way, I just started to learn this topic in the syllabus, which is equilibrium of rigid bodies. Questions from this topic often involves many unknowns, and I find myself having to form many simultaneous equations to solve them. So, I'm wondering if you have any tips on how to best approach these kind of equilibrium questions?

 

[Qwertywerty]

By the way, I just started to learn this topic in the syllabus, which is equilibrium of rigid bodies. Questions from this topic often involves many unknowns, and I find myself having to form many simultaneous equations to solve them. So, I'm wondering if you have any tips on how to best approach these kind of equilibrium questions?

Well , usually , the main part of the solution involves balancing torque .
Choose a point from which torque balancing is easy - in many cases , it is from where minimum torques are visible , in many it is not .

Force balancing always forms part of the equations .

Practice really helps .

 

[haruspex]

By the way, I just started to learn this topic in the syllabus, which is equilibrium of rigid bodies. Questions from this topic often involves many unknowns, and I find myself having to form many simultaneous equations to solve them. So, I'm wondering if you have any tips on how to best approach these kind of equilibrium questions?

See if the tip at the end of section 1 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/ helps.

 

[andrewkirk]

I still can't manage to express BC in terms of θ only. Here are my steps:

You were almost there. You had derived two equations, involving only $y,\theta,BC$. Since $\theta$ is given, you have two unknowns and two equations, which you can solve.

Here are the two equations you produced:

4/3 = tan θ (1.5-y)/ y
BC =5/4 tan θ (1.5-y)

Now just make y the subject of the first equation, then substitute that into the second equation and you'll have BC expressed in terms of only $\theta$.

 

[haruspex]

There are three tricks here that, combined, allow the solution to be written down in one go.

First, the tip I linked to in post 20. There are two unknown forces that you don't need to know, the two normal forces. So take moments about the point where they intersect. This one equation must lead to the answer. The tricky part is finding the moment due to the tension.

Second, resolve the tension into its horizontal and vertical components, T cos theta and T sin theta.

Third, treat the rod+rope as the 'system'. This allows you to treat T as being applied at the bottom of the rope instead of where it is tied to the rod. This makes finding the moments due to the components of T trivial.

 

[toforfiltum]

First, the tip I linked to in post 20. There are two unknown forces that you don't need to know, the two normal forces. So take moments about the point where they intersect. This one equation must lead to the answer. The tricky part is finding the moment due to the tension.

To confirm because I'm not too sure, the moments of these two normal forces intersect at O? And if it's correct, is it true that the way to find the intersection of two moments is by finding the intersection of their two perpendicular distances, as in the diagram?
I don't get what your last two sentences here. If you said that by taking the method above, you will get one equation that answers the problem, why must we again find the moment due to tension? Or are you saying that taking the moments of the normal forces is better than finding moment due to tension?

 

[Qwertywerty]

To confirm because I'm not too sure, the moments of these two normal forces intersect at O.

No , at a point ⊥ to OA and ⊥ to OB .

I don't get what your last two sentences here. If you said that by taking the method above, you will get one equation that answers the problem, why must we again find the moment due to tension? Or are you saying that taking the moments of the normal forces is better than finding moment due to tension?

Haruspex has provided a good method . From the point he speaks about , only tension and gravity exerts a torque . The only problem in this solution is finding the proper value of tension's torque .

 

[haruspex]

I don't get what your last two sentences here. If you said that by taking the method above, you will get one equation that answers the problem, why must we again find the moment due to tension? Or are you saying that taking the moments of the normal forces is better than finding moment due to tension?

I think Qwertywerty has explained this already, but I'll answer too.
We're not finding that moment again. We have to find it in order to write down that one equation. Finding it is the only tricky part in writing the equation.

 

[andrewkirk]

See if the tip at the end of section 1 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/ helps.

In the last equation in Example 1 of that tip (at the end of Section 1), should the left hand side be halved to $\frac{W}{2}(2L-b)$ to reflect the fact that the weight of the section of the beam to the left of B acts halfway along that section?

Those diagrams are beautiful. How do you make them? Is it laborious?

 

[andrewkirk]

To confirm because I'm not too sure, the moments of these two normal forces intersect at O?

The forces being referred to are applied at A by the wall, directly to the right, and at B by the floor, straight up. So the point you want to use as the centre of rotation for torque calcs is at the intersection of a line that goes vertically up from B with a line that goes horizontally to the right from A. In other words, it's the fourth corner of a rectangle whose other three corners are A, O and B..

And if it's correct, is it true that the way to find the intersection of two moments is by finding the intersection of their two perpendicular distances, as in the diagram?

No it's the intersection of lines of force we are finding, as described above, not the intersection of moments - which I don't think is a meaningful concept.

I don't get what your last two sentences here. If you said that by taking the method above, you will get one equation that answers the problem, why must we again find the moment due to tension?

Once we take the new point, call it X, as the centre of rotation, we can ignore the forces from the wall and the floor because their lines of action pass through X, so they exert no torque about X. The forces we are left with are:

(1) the weight W of the ladder, applied vertically downwards at the ladder's midpoint; and
(2) the tension of the string, which is easiest to handle as haruspex suggests, by considering it as being applied at O. So this is a force acting in the direction of the string, of size T, at O.

The torques about X from (1) and (2) are calculated by resolving each one into components parallel and normal to XO. Each torque is the normal component multiplied by the distance from X to the point at which it is applied. By looking at the diagram you can see that these torques act in opposite directions to one another.

For the system to be stationary, the magnitudes of the torques from (1) and (2) must be equal. Setting them equal gives us the one equation we need to solve for the one unknown T, since we are given W, $\theta$ and the angle of the ladder.

This is a far neater solution than my one. Thanks haruspex.

 

[Qwertywerty]

In the last equation in Example 1 of that tip (at the end of Section 1), should the left hand side be halved to W2(2L−b)\frac{W}{2}(2L-b) to reflect the fact that the weight of the section of the beam to the left of B acts halfway along that section?

Actually , I think it should be W(L-b) .

Haruspex , could you confirm ?

 

[haruspex]

Actually , I think it should be W(L-b) .

Haruspex , could you confirm ?

Thanks, yes, I'll fix it.

 

[toforfiltum]

No , at a point ⊥ to OA and ⊥ to OB .
Haruspex has provided a good method . From the point he speaks about , only tension and gravity exerts a torque . The only problem in this solution is finding the proper value of tension's torque .

The forces being referred to are applied at A by the wall, directly to the right, and at B by the floor, straight up. So the point you want to use as the centre of rotation for torque calcs is at the intersection of a line that goes vertically up from B with a line that goes horizontally to the right from A. In other words, it's the fourth corner of a rectangle whose other three corners are A, O and B..
No it's the intersection of lines of force we are finding, as described above, not the intersection of moments - which I don't think is a meaningful concept.

Once we take the new point, call it X, as the centre of rotation, we can ignore the forces from the wall and the floor because their lines of action pass through X, so they exert no torque about X. The forces we are left with are:

(1) the weight W of the ladder, applied vertically downwards at the ladder's midpoint; and
(2) the tension of the string, which is easiest to handle as haruspex suggests, by considering it as being applied at O. So this is a force acting in the direction of the string, of size T, at O.

The torques about X from (1) and (2) are calculated by resolving each one into components parallel and normal to XO. Each torque is the normal component multiplied by the distance from X to the point at which it is applied. By looking at the diagram you can see that these torques act in opposite directions to one another.

For the system to be stationary, the magnitudes of the torques from (1) and (2) must be equal. Setting them equal gives us the one equation we need to solve for the one unknown T, since we are given W, $\theta$ and the angle of the ladder.

This is a far neater solution than my one. Thanks haruspex.

Ahh...thanks for writing such a detailed explanation for the solution. I understand it more clearly now, though there are some loose ends to be tied up.

 

[toforfiltum]

Third, treat the rod+rope as the 'system'. This allows you to treat T as being applied at the bottom of the rope instead of where it is tied to the rod. This makes finding the moments due to the components of T trivial.

I don't see how treating the rod+rope as a system allows me to treat T as being applied only at the bottom of the rope instead of where it is tied to the rod. In class, my teacher taught me to either resolve the components into their separate bodies, where the reaction forces among the two bodies in contact cannot be canceled out and must be considered. However, she taught me, that if treated as a whole system, the reaction forces cancel out. Though I know it is not exactly the same case as in this question, I don't see how tension only applies to the bottom of the rope, I really don't.

 

[toforfiltum]

No , at a point ⊥ to OA and ⊥ to OB ./QUOTE]

Isn't that at O?

 

[toforfiltum]

By the way, though I understand andrewkirk's explanation, I wonder if moment can be calculated around an imaginary point X?

 

[toforfiltum]

Haruspex has provided a good method . From the point he speaks about , only tension and gravity exerts a torque . The only problem in this solution is finding the proper value of tension's torque .

So Haruspex's method is different than yours, is it not? Your method takes moments about O right?

 

[Qwertywerty]

So Haruspex's method is different than yours, is it not? Your method takes moments about O right?

Yes - his does not need you to take into account the normal forces . However you need to be careful while balancing the torques .

Isn't that at O?

O isn't perpendicular to the lines it lies on ( OA and OB ) .