# Weight & Tension Helo 2 Tension

• d1343
In summary, In this conversation, the individual is asking for help with a statics problem and specifically how to find the weight in order to solve for the sum of forces in the y direction. They are given a picture and the relevant equations, but are struggling to understand how to apply them to find the weight. The expert advises them to resolve the tensions into their X and Y components and to be careful as one of the tensions may not be critical to holding up the weight. The expert also suggests finding the critical constraint and using that information to determine the maximum weight that can be supported. The final answer is approximately 259.81 lbs.
d1343

## Homework Statement

How to find Weight in this problem?
I am trying to do this statics problem
but I don't know how to find the weight
so I can solve for
Sum Fy = 0
here is the picture link

## Homework Equations

TI
sum of Fx = 0
480 cos(30) - 450 = 0
-34

T1
sum of Fy = 0
480 sin(30) - W = 0
how do I find the weight to substract it

## The Attempt at a Solution

I don't know I am like really stuck

Last edited:
Resolve the tensions into their X and Y components.

The system is static, so the sum of the X forces must equal 0 and sum of the Y as well.

Be careful. One of the tensions may not be critical to holding up the weight.

wht the equation I wrote is not right?
Sum of the Force in X = 480 cos(30) - 450 = 0
I get = 415 - 450
= -34
Sum of the Force in y = 480 sin (30) - W = 0
240 -w = 0

I how do I solve for weight
the answer is about 259.81 lb
& there is also for t2

ΣFx = 0; 480cos(30) − 450 = 0

ΣFy = 0; 480sin(30) − W = 0

(F ab2 W2) = Find (FAB W)

W2 = 240.00 lb
W= min (W1 W2) = W 240.00 lb

You have 2 tensions to worry about.

Like the weakest link - that's where it will break right?

First figure out which is the critical constraint.

If T1 = 450 max, what does that translate to for the chain that can withstand 480? Will the 450 break if it's 480, or will the 480 break if it's 450?

Then you are armed with what you need to know, because from your equations, you can see that whatever the T1, the weight it can support is 1/2 T1.

## 1. What is the purpose of a Weight & Tension Helo 2 Tension?

A Weight & Tension Helo 2 Tension is a device that measures the weight and tension of an object or material. It is commonly used in industries such as construction, manufacturing, and transportation to ensure the proper handling and distribution of weight and tension.

## 2. How does a Weight & Tension Helo 2 Tension work?

The device works by using sensors to measure the weight and tension of an object. These sensors are connected to a display unit which shows the readings in real-time. The device may also include a data logging feature to record and analyze the weight and tension over time.

## 3. What are the benefits of using a Weight & Tension Helo 2 Tension?

There are several benefits to using a Weight & Tension Helo 2 Tension. It allows for accurate and precise measurements, which can help prevent damage to materials and equipment. It also improves safety by ensuring proper weight and tension distribution, reducing the risk of accidents. Additionally, it can help with quality control and efficiency in production processes.

## 4. Is a Weight & Tension Helo 2 Tension easy to use?

Yes, a Weight & Tension Helo 2 Tension is designed to be user-friendly and easy to operate. Most devices come with a user manual that provides step-by-step instructions on how to use it. Training may be required for first-time users, but once familiar with the device, it is relatively simple to use.

## 5. Are there different types of Weight & Tension Helo 2 Tension available?

Yes, there are various types of Weight & Tension Helo 2 Tension available, depending on the specific needs and requirements of the industry or application. Some devices may be handheld, while others may be mounted or integrated into equipment. There are also different weight and tension capacities available, so it is essential to choose the right device for the intended use.

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