Finding the point of equilibrium

[Augustine Duran]

Homework Statement

As a stunt coordinator with a background in physics, you have been asked to determine whether Man can make it across the river and whether the rope will hold his weight. Man is rather small, with a mass of only about 48.7 kg, including his equipment. The crossing distance is roughly W = 8.7 m and the rope is tied H = 1.7 m lower on the opposing the bank. Judging by the sag in the rope, its length is about 11.5 m. Calculate the tension in the rope once man reaches a point of equilibrium. Assume the rope does not stretch.

The Attempt at a Solution

I'm pretty sure I can find the tension of the rope once I find where the equilibrium point is, thing is I'm not sure HOW to find the equilibrium point. All i know is that the equilibrium point should have the same angle on both sides.

 

[andrewkirk]

The idea is to assign symbols to the various quantities of interest - lengths, angles, forces - then use those in a series of equations to describe the equilibrium point.

I assume, although they do not say this, that they want you to assume that at the equilibrium point the rope will be an asymmetric V shape, with the two sides being straight. In practice there would still be slight curves in either side of the rope, but modelling them is complex, so I think you're probably expected to ignore that.

Assuming that, what equations can you write that ensure equilibrium - ie net force on the man is zero?

EDIT: a potentially easier way to do this is to note that equilibrium will be where the man is at the lowest point, so all you need to do is maximise the distance below one of the tether points that bottom of the V can be, subject to the rope being of the specified length and connecting to both tether points. This approach is an 'energy' approach rather than a 'net force / free body diagram' approach.

 

[Augustine Duran]

So if the system is in equilibrium then the sum of the forces should equal zero
Ill label the left side T1 and the right side T2

Σ_x : -T1 cosθ + T2 cosθ = 0
T2 cosθ = T1 cosθ
T2 = T1

Σ_y : T1 sinθ + T2 sinθ - mg = 0
T1 sinθ + T2 sinθ =Mg

EDIT: we have not covered energy yet

 

[haruspex]

i know is that the equilibrium point should have the same angle on both sides

Right, so to find the angle it's just geometry.
Draw a diagram of the equilibrium position. In terms of the angle, how far is the lowest point of the rope below each of the two ends?

 

[Augustine Duran]

Right, so to find the angle it's just geometry.
Draw a diagram of the equilibrium position. In terms of the angle, how far is the lowest point of the rope below each of the two ends?

Would that be the altitude?

in terms of the angle all i can think of is this: sinθ = y/z → z sinθ = y

 

[Augustine Duran]

Also here's a diagram i drew, i imagine the rope would look like something like this at equilibrium.

 

[haruspex]

Also here's a diagram i drew, i imagine the rope would look like something like this at equilibrium.
View attachment 106753

Would that be the altitude

Yes.
It helps to add labels and construction lines. In you diagram, call the tops of the posts (same height) A (left) and B; the rope attaches at points A and C; the lowest point of the rope is D. Draw a vertical up from D to meet AB at E. Draw a horizontal from C to meet DE at F.
Now everything is reduced to rectangles and triangles with known angles. Using AED, what is length DE? Using CDF what is length DF?

 

[Augustine Duran]

so i redrew the diagram

problem is i don't understand how you can find the length of DE without knowing the length of AE. If for some reason you want me to use the length "6.2m", how would i know that's the equilibrium point? I probably should i said this in my original post but that length "6.2m" is used in the 2nd part of the question asking if the man can make it across the river and onto land.

 

[haruspex]

so i redrew the diagram

View attachment 106765

problem is i don't understand how you can find the length of DE without knowing the length of AE. If for some reason you want me to use the length "6.2m", how would i know that's the equilibrium point? I probably should i said this in my original post but that length "6.2m" is used in the 2nd part of the question asking if the man can make it across the river and onto land.

see what equations you can come up with to relate the various lengths. You could start by letting AE be x and see what expressions you can come up with for other lengths using x and the unknown angle to the horizontal. You should get enough equations to solve for x and the angle.
.

 

[Augustine Duran]

would

x tanθ = DE
and
DF tanθ =FC

be a good place to start? I'm kinda going in loops here.

 

[haruspex]

x tanθ = DE

Yes

DF tanθ =FC

Not quite.

You should also be able to write down simple relationships between DE and DF and between x and FC using the given lengths.
There is another given length that you need to make use of.

 

[Augustine Duran]

For DE and DF
We know the length of BC, which is the same length as EF
So...
DE-EF=DF

For x and FC
We know the length of EB is going to be the same length as FC, and AB is given
so can we say
AB-FC= x?

 

[haruspex]

DE-EF=DF

Yes, and you know EF.

AB-FC= x?

Yes, and you know AB.
What given length have you not got into an equation yet?

 

[Augustine Duran]

the rope?
would it be legal to add two lengths even though they meet at a corner?
so would it be...
AD+DC=AC

 

[haruspex]

the rope?
would it be legal to add two lengths even though they meet at a corner?
so would it be...
AD+DC=AC

No, "AC" would mean the straight line connecting them. But AD+DC= length of rope.

 

[Augustine Duran]

So are these the only equations i need to solve this?
DE-EF=DF
AB-FC=x
X tanθ=DE
AD+DC= rope
If so I am curious as to how i can incorporate the rope length since its not a straight line

 

[haruspex]

So are these the only equations i need to solve this?
DE-EF=DF
AB-FC=x
X tanθ=DE
AD+DC= rope
If so I am curious as to how i can incorporate the rope length since its not a straight line

Relate AD to x and theta. Similarly DC.