Moments Question (ENGAA 2019) -- Cable and Winch Lifting a Drawbridge

[TomK]

Homework Statement

Q18 (Section 2), ENGAA 2019

Relevant Equations

Moment equation.

The correct answer is 'B'.

I would like some clarification, in regard to the forces that exist in this moments problem.

General Queries:

1) Is there friction acting on the hinge in a downward direction (parallel to the wall)? Does the ramp being "smoothly hinged" to the wall mean there is no friction acting on the hinge?

If friction did exist, would there be a reaction force (since: friction = mu x reaction), acting perpendicular to the wall, from where the hinge is in-contact? Would the direction of said-reaction force change, depending on how the ramp is angled?

2) It is stated that the string is "light" (of negligible mass) and "inextensible"? If the string could extend, how would that affect the way you solve the problem? Wouldn't you just find the moments as normal? Would you have to do energy calculations, due to elastic potential energy in the string?

3) Where the string is fixed to the wall (at the top), does a reaction force exist, acting perpendicular to the wall, from where the string is in-contact?

4) Why does it matter that the ramp is lowered "at constant speed"? Would it affect the tension in the string?

Worked Solution Queries:

I looked at this worked solution (http://www.engineeringadmissionsassessment.com/2019-solutions.html) - scroll-down to the bottom of the page to see working for Q18, Section 2.

It looks like they start solving the problem when the ramp is fully-horizontal, when the angle between the wall and ramp is 90 degrees.

How can they say that the angle between the string and ramp is 'theta/2'? Doesn't this assume it's an isosceles triangle, where the 'distance between the winch and hinge' = 'length of ramp'?

I have a lot of confusion over this question and how to start it.

 

[etotheipi]

Is there friction acting on the hinge in a downward direction (parallel to the wall)? Does the ramp being "smoothly hinged" to the wall mean there is no friction acting on the hinge?

The hinge is assumed to be frictionless, i.e. it exerts no frictional force, or couple moment, on the ramp. A smooth hinge can only exert a reaction force, which may have both vertical and horizontal components [N.B. this reaction force is not necessarily parallel to the rod, and in most cases isn't!] . If you take the hinge as your origin of coordinates in this problem, these horizontal and vertical reaction forces can be ignored for purposes of balancing torques [since their torques about the hinge will evidently be zero!].

It is stated that the string is "light" (of negligible mass) and "inextensible"? If the string could extend, how would that affect the way you solve the problem? Wouldn't you just find the moments as normal? Would you have to do energy calculations, due to elastic potential energy in the string?

A string is special in that, in mechanics, it can transmit any tensile force required to fulfil the constraints of the problem, without any extension. If the problem instead used a spring, it would likely be specified that it obey's Hooke's law, and now there is an explicit relationship between the angle of the rod, and the tension force in the spring. The problem would change quite a bit.

Where the string is fixed to the wall (at the top), does a reaction force exist, acting perpendicular to the wall, from where the string is in-contact?

Yes, the wall exerts a contact force (with horizontal and vertical components) on the string. As it happens, this contact force is equal and opposite to the force exerted by the rod on the string [remember that the string is massless, so the net force on the string must be zero!]. You don't need to worry about this, however, if you define your system to be only the rod.

Why does it matter that the ramp is lowered "at constant speed"? Would it affect the tension in the string?

The condition of zero angular acceleration means that the net torque on the rod is zero, and you can balance torques. If angular acceleration was present, the net torque on the rod would no longer be zero!

How can they say that the angle between the string and ramp is 'theta/2'? Doesn't this assume it's an isosceles triangle, where the 'distance between the winch and hinge' = 'length of ramp'?

The triangle defined by the vertical rod, the rod at some angle $\theta$ and the line connecting the tips of these two [i.e. the cable as shown on the diagram] is an isosceles triangle, since the two larger sides both have length $l$.

 

[TomK]

The hinge is assumed to be frictionless, i.e. it exerts no frictional force, or couple moment, on the ramp. A smooth hinge can only exert a reaction force, which may have both vertical and horizontal components [N.B. this reaction force is not necessarily parallel to the rod, and in most cases isn't!] . If you take the hinge as your origin of coordinates in this problem, these horizontal and vertical reaction forces can be ignored for purposes of balancing torques [since their torques about the hinge will evidently be zero!].
A string is special in that, in mechanics, it can transmit any tensile force required to fulfil the constraints of the problem, without any extension. If the problem instead used a spring, it would likely be specified that it obey's Hooke's law, and now there is an explicit relationship between the angle of the rod, and the tension force in the spring. The problem would change quite a bit.
Yes, the wall exerts a contact force (with horizontal and vertical components) on the string. As it happens, this contact force is equal and opposite to the force exerted by the rod on the string [remember that the string is massless, so the net force on the string must be zero!]. You don't need to worry about this, however, if you define your system to be only the rod.
The condition of zero angular acceleration means that the net torque on the rod is zero, and you can balance torques. If angular acceleration was present, the net torque on the rod would no longer be zero!
The triangle defined by the vertical rod, the rod at some angle $\theta$ and the line connecting the tips of these two [i.e. the cable as shown on the diagram] is an isosceles triangle, since the two larger sides both have length $l$.

Thank you for the information.

I see, so the hinge (if not smooth) would exert friction on the ramp, not the wall it's attached to? I thought reaction force happened when two objects/surfaces are in-contact.

If friction requires a reaction force to be present, what causes a smooth hinge to have a reaction force?

Is the string being massless the reason why I don't need to consider the reaction force between the wall and string?

Also, how do I know for sure that there are two sides of equal length, so that two of the angles will be equal (as per the properties of an isosceles triangle)? How can I prove that 'distance between the winch and hinge' = 'length of ramp'?

 

[etotheipi]

I see, so the hinge (if not smooth) would exert friction on the ramp, not the wall it's attached to? I thought reaction force happened when two objects/surfaces are in-contact.

For all intents and purposes the hinge can be considered part of the wall [i.e. it's rigidly attached to the wall]. Of course there will be internal forces between the hinge and the wall, but you shouldn't worry about them. If the hinge is said to be frictional, then you can take that to mean that the hinge exerts a resistive couple moment on the ramp.

If friction requires a reaction force to be present, what causes a smooth hinge to have a reaction force?

A reaction, or contact, force exerted by one body on the other body is usually expressed as the sum of two components: friction [the tangential component], and a normal reaction force [the, well, normal component]. The normal component of the contact force can exist without the friction force, and vice versa!

Is the string being massless the reason why I don't need to consider the reaction force between the wall and string?

No, it's because you're taking the system to be just the ramp. That means you only need to consider the external forces acting on the ramp, not anything else.

Also, how do I know for sure that there are two sides of equal length, so that two of the angles will be equal (as per the properties of an isosceles triangle)? How can I prove that 'distance between the winch and hinge' = 'length of ramp'?

The tip of the ramp traces out a circle whose radius is the length of the ramp!

By the way, good luck for the ENGAA !

 

[Lnewqban]

Please, see:
https://en.m.wikipedia.org/wiki/Structural_support

The mechanism of the problem works pretty much like your arm.
Consider the humerus like the wall, the elbow joint like the hinge, the radius and ulna like the bridge and the biceps like the strings.

A hinge with high moment resistance would remove tension from the strings.
In the extreme case, the bridge could be half open while there is no tension, because the resistive moment due to high rate of friction to rotation would balance the moment induced by the weight of the bridge.

In that extreme case, the connection of the hinge to the wall would be fully supporting the weight of the bridge (linear reaction force) and the induced moment (rotational reaction force).

In similar way, if you try to lose a very tight bolt from an engine, you wil be excerting both, a torque or moment and a linear force on the head of the bolt via your wrench.

We could consider that bolt to be a hinge with huge frictional resistance to initial rotation to be unscrewed.

 

[TomK]

For all intents and purposes the hinge can be considered part of the wall [i.e. it's rigidly attached to the wall]. Of course there will be internal forces between the hinge and the wall, but you shouldn't worry about them. If the hinge is said to be frictional, then you can take that to mean that the hinge exerts a resistive couple moment on the ramp.
A reaction, or contact, force exerted by one body on the other body is usually expressed as the sum of two components: friction [the tangential component], and a normal reaction force [the, well, normal component]. The normal component of the contact force can exist without the friction force, and vice versa!
No, it's because you're taking the system to be just the ramp. That means you only need to consider the external forces acting on the ramp, not anything else.
The tip of the ramp traces out a circle whose radius is the length of the ramp!

By the way, good luck for the ENGAA !

Thank you for your help and patience. I've been practicing these questions for potential at-interview assessments, but I appreciate the wishes. The 2019 past paper is finished, but I believe this forum deserves a lot of the credit.

 

[etotheipi]

Thank you for your help and patience. I've been practicing these questions for potential at-interview assessments, but I appreciate the wishes. The 2019 past paper is finished, but I believe this forum deserves a lot of the credit.

I hope the unprecedented circumstances aren't causing too much of a headache . Feel free to PM me if you have questions or concerns!

 

[TomK]

I hope the unprecedented circumstances aren't causing too much of a headache . Feel free to PM me if you have questions or concerns!

Just to clarify, reaction force can exist with just one (not necessarily both) out of these components: 'friction' and the 'normal reaction force'?

Also, why is it we don't know the direction of the reaction force acting on the hinge? I get why you can ignore this when solving this question, since you take moments about the hinge. Isn't the reaction force just perpendicular to what it's in-contact with, or is that just the 'normal component'? It's just, at this stage, I haven't seen a reaction force that wasn't perpendicular to a surface that exerts it in a question. Is the reaction direction not known because the hinge is attached to two surfaces, the wall and ramp?

What did you mean by a 'resistive couple moment'? I know what a couple is, but what would that look like, since the hinge is on the end of the ramp?

 

[etotheipi]

Just to clarify, reaction force can exist with just one (not necessarily both) out of these components: 'friction' and the 'normal reaction force'?

That's right, yes.

Also, why is it we don't know the direction of the reaction force acting on the hinge? I get why you can ignore this when solving this question, since you take moments about the hinge. Isn't the reaction force just perpendicular to what it's in-contact with, or is that just the 'normal component'? It's just, at this stage, I haven't seen a reaction force that wasn't perpendicular to a surface that exerts it in a question. Is the reaction direction not known because the hinge is attached to two surfaces, the wall and ramp?

It's a little tricky. The normal contact force is always normal to the interface of two materials in contact, yes. But the interface at the hinge can be hard to picture.

You can think of a hinge as a cylindrical pin inside of a circular ring [which has a very slightly larger radius than the pin]. If you press the cylindrical pin against different sides of the circular ring, hopefully you can see that it's possible for the contact force between them to be in any direction, since you can find every possible orientation of the two interfaces.

That's why a hinge can transmit a reaction force in any possible direction, so you always need to leave two degrees of freedom available in your equations relating to the hinge, e.g. perhaps an unknown horizontal and vertical component, or a magnitude and an angle, etc.

In general, the different types of supports you'll encounter in mechanics, and the reactions they can provide, is summarised here:

As said before, it's very important to keep in mind that the reaction force provided by an external/internal pin or a fixed support can potentially be in any direction, and has no prior relation to the orientation of the beams to which the hinge is connected.

What did you mean by a 'resistive couple moment'? I know what a couple is, but what would that look like, since the hinge is on the end of the ramp?

A couple is just a pair of antiparallel forces of $\vec{F}$ and $-\vec{F}$ applied to a body in such a way that their lines of action do not coincide. A couple is special, since the torque of a couple is independent of the choice of origin. Furthermore, the vector sum of these forces is zero, so they solely contribute to the rotational dynamics and do not contribute to linear acceleration of the centre of mass.

If the hinge is not well oiled or it's a bit rusty, when you try to rotate the bar, friction at the hinge will exert an effective couple moment on the bar in the opposite sense to the torque which you're applying. In reality, you can think of this being made up of many many small tangential friction forces, however these can be replaced equivalently by a couple.

 

[TomK]

That's right, yes.
It's a little tricky. The normal contact force is always normal to the interface of two materials in contact, yes. But the interface at the hinge can be hard to picture.

You can think of a hinge as a cylindrical pin inside of a circular ring [which has a very slightly larger radius than the pin]. If you press the cylindrical pin against different sides of the circular ring, hopefully you can see that it's possible for the contact force between them to be in any direction, since you can find every possible orientation of the two interfaces.

That's why a hinge can transmit a reaction force in any possible direction, so you always need to leave two degrees of freedom available in your equations relating to the hinge, e.g. perhaps an unknown horizontal and vertical component, or a magnitude and an angle, etc.

In general, the different types of supports you'll encounter in mechanics, and the reactions they can provide, is summarised here:

View attachment 271293

As said before, it's very important to keep in mind that the reaction force provided by an external/internal pin or a fixed support can potentially be in any direction, and has no prior relation to the orientation of the beams to which the hinge is connected.
A couple is just a pair of antiparallel forces of $\vec{F}$ and $-\vec{F}$ applied to a body in such a way that their lines of action do not coincide. A couple is special, since the torque of a couple is independent of the choice of origin. Furthermore, the vector sum of these forces is zero, so they solely contribute to the rotational dynamics and do not contribute to linear acceleration of the centre of mass.

If the hinge is not well oiled or it's a bit rusty, when you try to rotate the bar, friction at the hinge will exert an effective couple moment on the bar in the opposite sense to the torque which you're applying. In reality, you can think of this being made up of many many small tangential friction forces, however these can be replaced equivalently by a couple.

Thank you for explaining.

 

[TomK]

That's right, yes.
It's a little tricky. The normal contact force is always normal to the interface of two materials in contact, yes. But the interface at the hinge can be hard to picture.

You can think of a hinge as a cylindrical pin inside of a circular ring [which has a very slightly larger radius than the pin]. If you press the cylindrical pin against different sides of the circular ring, hopefully you can see that it's possible for the contact force between them to be in any direction, since you can find every possible orientation of the two interfaces.

That's why a hinge can transmit a reaction force in any possible direction, so you always need to leave two degrees of freedom available in your equations relating to the hinge, e.g. perhaps an unknown horizontal and vertical component, or a magnitude and an angle, etc.

In general, the different types of supports you'll encounter in mechanics, and the reactions they can provide, is summarised here:

View attachment 271293

As said before, it's very important to keep in mind that the reaction force provided by an external/internal pin or a fixed support can potentially be in any direction, and has no prior relation to the orientation of the beams to which the hinge is connected.
A couple is just a pair of antiparallel forces of $\vec{F}$ and $-\vec{F}$ applied to a body in such a way that their lines of action do not coincide. A couple is special, since the torque of a couple is independent of the choice of origin. Furthermore, the vector sum of these forces is zero, so they solely contribute to the rotational dynamics and do not contribute to linear acceleration of the centre of mass.

If the hinge is not well oiled or it's a bit rusty, when you try to rotate the bar, friction at the hinge will exert an effective couple moment on the bar in the opposite sense to the torque which you're applying. In reality, you can think of this being made up of many many small tangential friction forces, however these can be replaced equivalently by a couple.

I did this question a while ago, but I wanted to ask how you know that the maximum tension occurs when θ = 90 degrees.

I worked out the moments, and I got:

tension = Wcos(θ/2)

I'm thinking that I did it wrong.

We want cos(θ/2) to be as large as possible to maximise tension. Of course, θ can't be 0 degrees, otherwise there will not be tension. We know θ can only increase to 90 degrees in this question.

The correct answer requires θ = 90 degrees, so cos(θ/2) = cos(45). However, shouldn't you get a greater value by using a lower θ value? Please help.

 

[etotheipi]

I think you've made a small trigonometry mistake somewhere (e.g. maybe writing a $\sin$ when you should have written a $\cos$, or vice versa), it'd help if you show your working. About the hinge, the magnitude of the torque of the weight is$$\Gamma_{W} = (W\sin{\theta}) \times \frac{l}{2} = Wl \sin{\left(\frac{\theta}{2} \right)}\cos{\left(\frac{\theta}{2} \right)}$$whilst the magnitude of the torque of the tension force is$$\Gamma_{T} = T\sin{\left( \frac{\pi - \theta}{2} \right)} \times l = Tl \cos{\left(\frac{\theta}{2} \right)}$$Since the rod is moving with constant angular velocity, it's in rotational equilibrium and these torques are equal in magnitude, so in the interval we're interested in$$W \sin{\left( \frac{\theta}{2} \right)} = T$$Since $\sin{x}$ is monotonically increasing in $[0, \pi/2]$ the maximum tension must occur when $\theta = \pi/2$, i.e. when $\sin{\left( \frac{\theta}{2} \right)} = 1/\sqrt{2}$.

 

[TomK]

I think you've made a small trigonometry mistake somewhere (e.g. maybe writing a $\sin$ when you should have written a $\cos$, or vice versa), it'd help if you show your working. About the hinge, the magnitude of the torque of the weight is$$\Gamma_{W} = (W\sin{\theta}) \times \frac{l}{2} = Wl \sin{\left(\frac{\theta}{2} \right)}\cos{\left(\frac{\theta}{2} \right)}$$whilst the magnitude of the torque of the tension force is$$\Gamma_{T} = T\sin{\left( \frac{\pi - \theta}{2} \right)} \times l = Tl \cos{\left(\frac{\theta}{2} \right)}$$Since the rod is moving with constant angular velocity, it's in rotational equilibrium and these torques are equal in magnitude, so in the interval we're interested in$$W \sin{\left( \frac{\theta}{2} \right)} = T$$Since $\sin{x}$ is monotonically increasing in $[0, \pi/2]$ the maximum tension must occur when $\theta = \pi/2$, i.e. when $\sin{\left( \frac{\theta}{2} \right)} = 1/\sqrt{2}$.

I can't see where I've made a mistake.

I just used: moment = force x perpendicular distance (from hinge to direction of force)If you make a triangle connecting hinge and force going down from centre of ramp:

sin(θ) = O/H
sin(θ) = x1/(l/2)

x1 = lsin(θ)/2

CW moment = Wlsin(θ)/2If you make a triangle connecting hinge and string:

sin(θ/2) = x2/l
x2 = lsin(θ/2)

ACW moment = Tlsin(θ/2)

(equate moments): T = [Wsin(θ)] / [2sin(θ/2)]

double angle formula: sin(2θ) = 2sin(θ)cos(θ)

sin(θ) = 2sin(θ/2)cos(θ/2)

T = 2Wsin(θ/2)cos(θ/2) / 2sin(θ/2)

T = Wcos(θ/2)

 

[etotheipi]

If you make a triangle connecting hinge and string:

sin(θ/2) = x2/l
x2 = lsin(θ/2)

This is the mistake, the perpendicular distance is$$x_2 = l\cos{\left( \frac{\theta}{2} \right)}$$

 

[TomK]

This is the mistake, the perpendicular distance is$$x_2 = l\cos{\left( \frac{\theta}{2} \right)}$$

I don't understand why.

You have the angle θ/2 in the bottom-right corner (between ramp and string). The 90 degree angle is at the top corner. The triangle appears to also be isoceles.

If you use the angle by the string, you get x2 to be the opposite side and the ramp to be the hypotenuse side. Why not use 'sin'?

 

[etotheipi]

The angle between the dotted green line (whose length is the perpendicular distance) and the rod is $\theta / 2$. Then $x_2 = l \cos{\left( \frac{\theta}{2}\right)}$

 

[etotheipi]

You have the angle θ/2 in the bottom-right corner (between ramp and string). The 90 degree angle is at the top corner. The triangle appears to also be isoceles.

N.B. the angle between the ramp and the string is actually $(\pi - \theta)/2$, not $\theta/2$.

 

[TomK]

N.B. the angle between the ramp and the string is actually $(\pi - \theta)/2$, not $\theta/2$.

That's it. I confused θ with pi at the very beginning. Thank you.