Finding Forces and Masses in a Hanging Weight System

In summary: It is a number, but is it the weight in N ?In summary, the force in thread B is approximately 21.2 N and the mass of the weight is approximately 1.47 kg. The key to solving this problem is understanding that the lengths of the strings do not affect the forces, but the angles do. By analyzing the forces on the point where all three strings are connected, it is possible to determine the forces and mass involved.
  • #1
BearMan
3
0
Just started with physics, I am not very good.

Homework Statement


A weight with the mass m is hanging in one thread (C) that is connected to two other threads (A and B). Thread A is completely horizontal, while thread B is forming a angle of 45° with the horizontal plane. All threads are connected in the way shown in the image. The force in thread A is 15N.
My task is to figure out the force in thread B and the mass of the weight.
g77OjOs.png


Homework Equations





The Attempt at a Solution


I thought that a simple use of the Pythagorean theorem would give me the power in thread B; the triangle is 4 in height, 4 in weight. The A "contains" 15N of energy and is 6 units long, therefore i divided 15 by 6 and got 2.5.
15/6=2.5
(4*2.5)^2+(4*2.5)^2=200
√(200)≈14.14N=B
That is wrong, I am sure.

I am not really sure how to solve this and I do not want the numerical answers to the questions, but rather tips and methods to solving the task.
We were not really supposed to do this perticular assignment (our teacher told us it was "stuff we had not gone through yet") but I am curious to know how one should go about solving it.
 
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  • #2
You are given the force in thread A. What do you know about the force(s?) on the point where A, B and C are joined togethrer, if this point isn't moving ?
 
  • #3
Oh, and: Hello ursus and welcome to PF!

The grid is distracting you. A force on a string at one end is carried along to the other end, irrespective of the length of the string. You can check that with a spring (or a rubber band), a piece of (lightweight) string and a suitable weight.

The nice thing about strings is that they can only exercise a force along the (taut) string. (if not "strung" then no force at all :smile:).

Bottom line: the length of the various strings doesn't matter. The angles do. So if string A is pulling to the left and they tell you the force is 15 N, something else has to pull back with the same force -- otherwise there is no equilibrium and movement would result.

In your case, wirde C can't do that, because all it does is pull in the vertical direction. Leaves only B. B is pulling up and to the right with equal forces, hence the 45 degree angle.

Enough "hints" for now, go to work with this and either come with new questions, or with the solution ! Good luck.

In the mean time, if you think something qualifies as "relevant equation", be sure to post it !
 
  • #4
BvU said:
Oh, and: Hello ursus and welcome to PF!

The grid is distracting you. A force on a string at one end is carried along to the other end, irrespective of the length of the string. You can check that with a spring (or a rubber band), a piece of (lightweight) string and a suitable weight.

The nice thing about strings is that they can only exercise a force along the (taut) string. (if not "strung" then no force at all :smile:).

Bottom line: the length of the various strings doesn't matter. The angles do. So if string A is pulling to the left and they tell you the force is 15 N, something else has to pull back with the same force -- otherwise there is no equilibrium and movement would result.

In your case, wirde C can't do that, because all it does is pull in the vertical direction. Leaves only B. B is pulling up and to the right with equal forces, hence the 45 degree angle.

Enough "hints" for now, go to work with this and either come with new questions, or with the solution ! Good luck.

In the mean time, if you think something qualifies as "relevant equation", be sure to post it !

Thanks alot. Did not expect that it would require almost no calculations. B is the only thing holding C from going to the left and since there is no motion, B must pull to the right with a force of 15 N. And also, since it is tilted at 45° then the force up must be equally as strong.
The weight:
15*9.82=1.47kg

And the B thread:
√(15^2+15^2)≈21.2 N
 
  • #5
Looks good, except the weight. Could you check that ?
 

What is "Two string with one weight"?

"Two string with one weight" refers to a scientific experiment in which two strings of equal length are attached to a weight and suspended from the ceiling. This experiment is used to demonstrate the principles of balance and tension in physics.

What materials are needed for the experiment?

To perform the "Two string with one weight" experiment, you will need two equal lengths of string, a weight (such as a small ball or block), and a sturdy ceiling or support to hang the strings from.

What is the purpose of the experiment?

The purpose of the "Two string with one weight" experiment is to demonstrate the principles of balance and tension in physics. By adjusting the length and tension of the strings, you can observe how the weight is affected and how the forces of tension act on the strings.

What are the variables in the experiment?

The variables in the "Two string with one weight" experiment include the length and tension of the strings, as well as the weight of the object. These variables can be adjusted to observe how they affect the balance and tension in the system.

What are the potential applications of this experiment?

The principles demonstrated in the "Two string with one weight" experiment can be applied to various real-world scenarios, such as designing bridges and other structures, understanding the mechanics of objects suspended by cables or ropes, and even studying the behavior of molecules in chemistry and biology.

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