# Tension in two ropes with mass

• rsrunner
Interesting excercise to work out the minimum T2 for T1 at 20 degrees and T2 attached so that the bar is in equilibrium (haru post #20). Turns out to be easy: 70 degrees. With the same 4022 for T2 !In summary, a 1200 kg steel beam is supported by two ropes, one at 20 degrees off of the vertical and the other at 30 degrees off of the vertical. The weight of the beam is equal to the sum of the tensions of the ropes. Using the equations F=ma and F=mg, it can be determined that the weight of the beam is -11,760 N. To find the tension on rope two, the beam is in

## Homework Statement

A 1200 kg steel beam is supported by two ropes. Rope one is 20 degrees off of the vertical, and rope two is 30 degrees off of the vertical. What is the tension on rope two? (Answer should look like '2300 N'.)

## Homework Equations

F=ma
The weight of the mass is defined as the force of gravity on the object. therefore W=(1,200 kg)(-9.8 m/s squared). W= -11,760

## The Attempt at a Solution

?
W=(1,200 kg)(-9.8 m/s squared). W= -11,760

Now I am not sure what do to.[/B]

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rsrunner said:

## Homework Statement

A 1200 kg steel beam is supported by two ropes. Rope one is 20 degrees off of the vertical, and rope two is 30 degrees off of the vertical. What is the tension on rope two? (Answer should look like '2300 N'.)

## Homework Equations

F=ma
The weight of the mass is defined as the force of gravity on the object. therefore W=(1,200 kg)(-9.8 m/s squared). W= -11,760

## The Attempt at a Solution

?
W=(1,200 kg)(-9.8 m/s squared). W= -11,760

Now I am not sure what do to.[/B]
The beam is in equilibrium. What does it mean on the sum of forces acting on it? Draw the FBD.

ehild said:
The beam is in equilibrium. What does it mean on the sum of forces acting on it? Draw the FBD.
That the forces( or tensions of the ropes) are equal to each other?

rsrunner said:
That the forces( or tensions of the ropes) are equal to each other?
No!

The W of the object is equal to the sum of the tensions of the ropes.

W= T1 + T2

I know W.

But how am I to find the T1 or T2? I don't understand the vector/ breaking into two equations.

Tension in Rope 1
cos 20= (W/T), therefore T1=W/cos 20

Tension in Rope 2
cos 30=(W/T), therefore T1=W/cos 30

rsrunner said:
No!

The W of the object is equal to the sum of the tensions of the ropes.

No!

The beam isn't accelerating in any direction so the sum of the forces acting an any direction is zero. eg The horizontal components of all forces sum to zero. The vertical components of all forces sum to zero. Write these two simultaneous equations.

T1 + T2 = 11760

T1x + T2x = 0

T1y + T2y = 0

right?

In which direction is W working ?

vertically

CWatters said:
The vertical components of all forces sum to zero.
Means T1y + T2y + W = 0

• rsrunner
I would write the two equations as:

T1y + T2y - W = 0
and
T1x - T2x = 0

However I don't get the answer T2 = 2300N.

• rsrunner

2300 N can't be right: with rope 1 at 20 degrees, there's a minimum T2 of over 4000 N

Pity there isn't a drawing. I initially imagined a beam supported from the ends like this \___/ but I think that would imply a non uniform beam (otherwise the angles would be the same). So I assume the beam must be hanging like this __\/__ Can the rsrunner please confirm?

CWatters said:
Pity there isn't a drawing. I initially imagined a beam supported from the ends like this \___/ but I think that would imply a non uniform beam (otherwise the angles would be the same). So I assume the beam must be hanging like this __\/__ Can the rsrunner please confirm?
Would you hang a heavy beam at its centre? Supposing the beam is horizontal the setup migh look like in the figure. The problem did not state that the beam was uniform. Last edited:
CWatters said:
I would write the two equations as:

T1y + T2y - W = 0
and
T1x - T2x = 0

However I don't get the answer T2 = 2300N.
Answer should look like '2300 N'.
does not mean that 2300 N is the answer. It means that you have to give the result in that form, perhaps, with two signifiacant digits.

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OP doesn't say ropes are attached at the ends

BvU said:
OP doesn't say ropes are attached at the ends

True... So the beam can be uniform, but it does not matter.

Wouldn't a very non uniform beam allow T2 to be 2300N?

##11760 \; \sin(20^\circ) = 4022##

CWatters said:
Wouldn't a very non uniform beam allow T2 to be 2300N?
I don't see how. You can have the beam at an angle, non-uniform density, ropes attached anywhere - you still have two linear force equations containing only two unknowns, the tensions. The torque equation then puts constraints on points of attachment, etc.

Having thought about it some more I agree.

BvU said:
##11760 \; \sin(20^\circ) = 4022##

??

CWatters said:
??
That was to support my post #12: with one rope at 20 degrees and all the weight on that one, T2 needs to compensate the horizontal component only. Still, that's over 4000 N. You'd need a very long rope to get to that minimum for a horizontal bar. Gets better if bar is at an angle and the light end almost touches the ceiling...

I thought that was a global minimum/extreme case. Not so:

Interesting excercise to work out the minimum T2 for T1 at 20 degrees and T2 attached so that the bar is in equilibrium (haru post #20). Turns out to be easy: 70 degrees. With the same 4022 for T2 !