MHB Find Tension in Horizontal Rope DE | Cylinder Mass & Angle θ

[Ciaran]

Hi there,

I have a question I'm unable to get properly started on:

A cylinder of radius R and mass m is lodged between cross pieces that make
an angle θ with each other, as shown in the diagram below. The cross pieces
of negligible mass are connected in point C, with lengths AC = BC = 2R and
CD = CE = 3R. Determine the tension in the horizontal rope DE. You may
assume smooth floor.

Any help would be much appreciated!
View attachment 4083

 

[I like Serena]

Hi there,

I have a question I'm unable to get properly started on:

A cylinder of radius R and mass m is lodged between cross pieces that make
an angle θ with each other, as shown in the diagram below. The cross pieces
of negligible mass are connected in point C, with lengths AC = BC = 2R and
CD = CE = 3R. Determine the tension in the horizontal rope DE. You may
assume smooth floor.

Any help would be much appreciated!

Hi Ciaran!

Here's the procedure to tackle any problem in statics.

1. Make an inventory of the external forces.
2. Draw a Free Body Diagram (FBD). That is a diagram of the object with only the external forces.
3. Apply the equilibrium equations of statics: $\sum F_x=0, \sum F_y=0, \sum T = 0$.
4. Make an inventory of the internal forces on each part.
5. Draw an FBD of each part.
6. Apply the equilibrium equations of statics on each part: $\sum F_x=0, \sum F_y=0, \sum T = 0$.

Which external forces can you identify?
Can you already say something about their sizes?

 

[Ciaran]

Hello again! I know the weight of the system will be the weight of the mass, m and that a reaction force will be split evenly and applied to either bit touching the ground. So it obviously isn't moving in the x or y direction. It's internally, I'm not sure how to deal with the contact forces between the mass and the cross rails.

 

[I like Serena]

Hello again! I know the weight of the system will be the weight of the mass, m and that a reaction force will be split evenly and applied to either bit touching the ground. So it obviously isn't moving in the x or y direction. It's internally, I'm not sure how to deal with the contact forces between the mass and the cross rails.

Good.

Let's give them a name. Let's say G is the force of gravity on the cylinder.
Then at A and B we will have an upward force of G/2 each, yes?
At A and B the horizontal component of force is 0, since the floor gives no friction.
That takes care of steps 1 and 3 and we've been sort of skipping step 2.

Where the rail touches the cylinder, there will be a normal force, perpendicular to the rail.
There won't be friction (parallel to the rail) because the cylinder cannot move/slide along the rail.

If we only look at the cylinder, which forces act on it?
What are their sizes?

 

[Ciaran]

The only force on the cylinder is G vertically downwards so this must be countered by the two forces from the cross rails, radially?

 

[I like Serena]

The only force on the cylinder is G vertically downwards so this must be countered by the two forces from the cross rails, radially?

Yes. Let's call them N, which is short for normal force.

 

[Ciaran]

Then the components that are horizontal will provide the tension?

 

[I like Serena]

Then the components that are horizontal will provide the tension?

No, we're not there yet.
The tension force does not work on the cylinder, it works on the rail.
To find the tension, first we need the normal force.

I suggest to try and draw an FBD for the cylinder, and also one for the rail.
We'll need them.

 

[Ciaran]

My FBD for the cylinder has 3 forces- gravity vertically downwards, and 2 radial forces 'inwards'. And for the rail there are 2 forces, each acting 'outward' due to each side of the rails being pulled down by the mass?

 

[I like Serena]

My FBD for the cylinder has 3 forces- gravity vertically downwards, and 2 radial forces 'inwards'.

Good!

And for the rail there are 2 forces, each acting 'outward' due to each side of the rails being pulled down by the mass?

On one rail there is indeed 1 outward acting force by the cylinder.
There are 3 more forces...

 

[Ciaran]

The weight of the rail? But it's considered negligible. And contact forces from where it touches two other rails.

 

[I like Serena]

The weight of the rail? But it's considered negligible. And contact forces from where it touches two other rails.

No, not the weight of the rail.

Each rail has 4 forces acting on it:
1. The tension of the rope. Let's call it T.
2. The normal force of the cylinder, which I've already called N.
3. The force of the other rail where it is connected in point C. Let's call it R (for rail).
4. The upward force of the floor, which is G/2, since it's the only external force to counter the weight G of the cylinder.

 

[Ciaran]

I understand that, yes. I'm just not sure about putting it all together. Have you an expression for the tension? I have one but it seems rather complicated and I'm not sure my method was correct. Does the answer involve secant and tangents of half of theta?

 

[I like Serena]

I understand that, yes. I'm just not sure about putting it all together. Have you an expression for the tension? I have one but it seems rather complicated and I'm not sure my method was correct. Does the answer involve secant and tangents of half of theta?

Yes, it involves a couple of trigonometric functions of half of theta.

 

[Ciaran]

It's not T= (1/6R)(R_N sec(theta/2))+ (1/3)Gtan(theta/2) is it? Where R_N is the radial force, R the radius of the cylinder and G as discussed before. Apologies for the typing- I'm not exactly an expert in using the software for equations!

 

[I like Serena]

It's not T= (1/6R)(R_N sec(theta/2))+ (1/3)Gtan(theta/2) is it? Where R_N is the radial force, R the radius of the cylinder and G as discussed before. Apologies for the typing- I'm not exactly an expert in using the software for equations!

I have something similar but not the same, and in particular I evaluated $R_N$.

As for "using the software for equation", just put a dollar on each side of an equation. That should do the trick.
And if you want to be a bit more fancy, put a backslash (\) in front of some of the symbols like \theta and \tan.