# Tension in two ropes with mass

## 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]

Last edited:

ehild
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## 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.

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?

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

CWatters
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Gold Member
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?

BvU
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In which direction is W working ?

vertically

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

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

CWatters
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Gold Member
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?

ehild
Homework Helper
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.

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ehild
Homework Helper
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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BvU
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OP doesn't say ropes are attached at the ends

ehild
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OP doesn't say ropes are attached at the ends

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

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

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

haruspex
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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.

CWatters
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Gold Member
Having thought about it some more I agree.

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

??

BvU