# Mass hanging by two ropes, find rope tensions.

• aitee
In summary, the weight of the object hanging from the ceiling by two ropes is -13720N. The first rope is 20 degrees to the left of the steel attach point, and the second rope is 30 degrees to the right of the attach point. The tension in the ropes is holding the object in the air. Newton's second law states that the force of gravity on the object is equal to the weight of the object. The equation for calculating the weight of the object is F = ma, where m is the weight in kilograms and a is the gravitational acceleration. The weight of the object is -13720N, and the tension in the ropes is holding the object in the air.
aitee

## Homework Statement

There is a steel beam of 1400kg hanging from the ceiling by two ropes. The first rope (R1) is 20 degrees to the left of the steel attach point, the second rope is 30 degrees to the right of the attach point.

Here is the diagram: http://img202.imageshack.us/img202/7654/06p29.jpg

## Homework Equations

The relevant equations would be Newton's second law, F = ma

Fx = F cos(theta)
Fy = F sin(theta)

This are all the relevant equations as far as I know.

## The Attempt at a Solution

First, I found W (weight, the force of gravity on the object) by simply using Newton's second law: F = ma where m = 1400kg and a = -9.8; with that I get W = -13720N

As the steel beam is not moving side to side, R1x = R2x Fnetx = 0

Because the tension is holding the steel beam in the air, the x-components of R1 and R2 must be equal to W. So: W = R1 cos 20 + R2 cos 30

Is this correct so far?

This is the part I get stuck. I know neither R1 or R2 and am not sure where to go.Any help would be appreciated so much.

Last edited by a moderator:
aitee said:

## Homework Statement

There is a steel beam of 1400kg hanging from the ceiling by two ropes. The first rope (R1) is 20 degrees to the left of the steel attach point, the second rope is 30 degrees to the right of the attach point.

Here is the diagram: http://img202.imageshack.us/img202/7654/06p29.jpg

## Homework Equations

The relevant equations would be Newton's second law, F = ma

Fx = F cos(theta)
Fy = F sin(theta)

This are all the relevant equations as far as I know.

## The Attempt at a Solution

First, I found W (weight, the force of gravity on the object) by simply using Newton's second law: F = ma where m = 1400kg and a = -9.8; with that I get W = -13720N

As the steel beam is not moving side to side, R1x = R2x Fnetx = 0

Because the tension is holding the steel beam in the air, the x-components of R1 and R2 must be equal to W. So: W = R1 cos 20 + R2 cos 30

Is this correct so far?
yes, but you seem to be considering the vertical axis as the x axis, which is OK, but uncoventional
This is the part I get stuck. I know neither R1 or R2 and am not sure where to go.

Any help would be appreciated so much.
Put R1x and R2x into their vector component form , just like you did in the vertical direction, and solve the 2 equations with 2 unknowns.

Last edited by a moderator:
Oh, I meant to be doing y.. Our teacher (she's confusing) said that because the angle is not along the x-axis and is on the other side of the vector that the y component would be cos instead of sin..

Sorry, I'm not trying to do it in an unconventional way, I want to do it the simplest and most straightforward way possible.

What I mean is, sin(30) == opposite/hypotenuse; so in the diagram (using a triangle) this would be Fy = F sin 30 == (F hyp)*(parallel to the x-axis opp) /hyp

Does this make sense? Like am I understanding why the component equation is what it is?

We use sin because is opp/hyp and when we multiply that by the vector magnitude(which is the hypotenuse) it results in opp; which is the component we were looking for in the first place?

Your teacher is correct..the y (vertical) component of Rope 2 is R2(cos 30) or R2(sin 60)...same result. Proceed as you were doing, just keep your components straight...and axes...

Thanks, I got it figured out now. I just had to get each component in the form Fx/y = F cos/sin(theta) and then set each equaling both W and zero and plug one into the other.

I'm not sure why I was so confused in the first place, hindsight--blah.

Thanks again.

## 1. How do you calculate the tension in each rope?

The tension in each rope can be calculated by using the equation T = (mg + ma)/2, where T is tension, m is mass, g is the acceleration due to gravity, and a is the acceleration of the hanging object.

## 2. What factors affect the tension in the ropes?

The tension in the ropes is affected by the mass of the hanging object, the acceleration of the object, and the angle of the ropes with respect to the horizontal.

## 3. Is the tension the same in both ropes?

In most cases, the tension in the ropes will not be the same. The rope with the steeper angle will have a higher tension, while the rope with the shallower angle will have a lower tension.

## 4. How do you find the angle between the ropes?

The angle between the ropes can be found by using the inverse tangent function, tan^-1 (m/a), where m is the mass and a is the acceleration of the hanging object.

## 5. Can the tension in the ropes ever be greater than the weight of the hanging object?

Yes, the tension in the ropes can be greater than the weight of the hanging object if there is an external force acting on the object, such as wind or another object pushing against it. In this case, the tension in the ropes will be equal to the weight of the object plus the force of the external force.

Replies
4
Views
345
Replies
7
Views
8K
Replies
1
Views
3K
Replies
22
Views
7K
Replies
4
Views
2K
Replies
5
Views
7K
Replies
13
Views
2K
Replies
5
Views
2K
Replies
5
Views
4K
Replies
9
Views
42K