Force on 2 cables with weight attatched using Theta....

Click For Summary

Homework Help Overview

The discussion revolves around calculating the forces on two cables supporting a weight, with angles θ_1 = 55° and θ_2 = 35°. Participants are analyzing the equilibrium of forces acting on the cables and questioning the application of gravitational force in their calculations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the equation F_1 + F_2 + F_g = 0 and the vector representations of the forces. There are questions regarding the correctness of the calculations and the interpretation of the weight of the block in different unit systems.

Discussion Status

Some participants have provided insights on potential errors in the calculations, particularly regarding the gravitational force and unit conversions. There is an ongoing exploration of the implications of using different units and the clarity of the problem statement.

Contextual Notes

Participants note the ambiguity in the problem statement regarding the forces acting on the cables and the weight of the block, which is given in pounds. There is a discussion about whether to incorporate gravitational acceleration in the calculations or to treat the weight as a force directly.

Unicow
Messages
14
Reaction score
0

Homework Statement


https://ibb.co/g6iKDQ
upload_2017-6-28_7-46-21.png


Calculate the magnitude of the force on cables 1 and 2 in the figure below if θ_1 = 55° and θ_2 = 35°.
(Round your answers to two decimal places.)

Homework Equations



Is my method incorrect or did I go wrong somewhere in my calculations??

The Attempt at a Solution


So what I did was I used the equation for F_1 + F_2 + F_g = 0.

F_1 = f_1 * <cos(125˚), sin(125˚)>
F_2 = f_2 * <cos(35˚), sin(35˚)>
F_g = 9.8 * (50) = 490<0, -1>

Then set F_1 + F_2 + F_g = <0, 0>

Moving to the system of equations
f_1*cos(125˚) + f_2*cos(35˚) = 0
f_1*sin(125˚) + f_2*sin(35˚) - 490 = 0

Solved for f_2 on first equation
f_2 = -f_1(cos(125˚)/cos(35˚))

plugged it into the 2nd equation
f_1*sin(125˚) + (-f_1(cos(125˚)/cos(35˚))sin(35˚) - 490 = 0

Solved for f_1 and got 502.46 rounded
and for f_2 I got 351.82.

After that, I plugged it into the F_1 and F_2 equations by
F_1 = 502.46<cos(125˚), sin(125˚)>
F_2 = 351.82<cos(35˚), sin(35˚)>

F_1 = <-288.2, 411.59>
F_2 = <288.2, 201.8>

Then got the magnitude with
Sqrt(288.2^2 + 411.59^2) = 502.46
Sqrt(288.2^2 + 201.8^2) = 351.83
 
Last edited by a moderator:
Physics news on Phys.org
Unicow said:

Homework Statement


https://ibb.co/g6iKDQ
I don't know how to make the pictures work...
Calculate the magnitude of the force on cables 1 and 2 in the figure below if θ_1 = 55° and θ_2 = 35°.
(Round your answers to two decimal places.)
Hello Unicow, :welcome:

Pictures go with copy/paste in windows. Don't know how to do it on a phone.
upload_2017-6-28_12-37-41.png

The problem statement is somewhat unfortunate: at each end of a cable equal but opposite forces are acting on the cable.
IMHO asking for the tensions in the cable is much clearer. Never mind.

I used the equation for F_1 + F_2 + F_g = 0
May be correct, but you don't tell us what they are...
F_1 = f_1 * <cos(125˚), sin(125˚)>
F_2 = f_2 * <cos(35˚), sin(35˚)>
F_g = 9.8 * (50) = 490<0, -1>
Again, you don't tell. I can guess, but that's not the idea!
Solved for f_1 and got 502.46 rounded
and for f_2 I got 351.82.
Did you check your answer to see if
f_1*sin(125˚) + f_2*sin(35˚) - 490 = 0
You do bother to continue finding ##\vec F_1, \vec F_2## (even though the exercise doesn't ask that) but fail to notice that ##411.59+201.8-490\ne 0##

And then you go on and calculate the magnitudes you had already !
 
Unicow said:

Homework Statement


https://ibb.co/g6iKDQ
View attachment 206199

Calculate the magnitude of the force on cables 1 and 2 in the figure below if θ_1 = 55° and θ_2 = 35°.
(Round your answers to two decimal places.)

Homework Equations



Is my method incorrect or did I go wrong somewhere in my calculations??

The Attempt at a Solution


So what I did was I used the equation for F_1 + F_2 + F_g = 0.

F_1 = f_1 * <cos(125˚), sin(125˚)>
F_2 = f_2 * <cos(35˚), sin(35˚)>
F_g = 9.8 * (50) = 490<0, -1>

Whoa! The question said the block weighs 50 lb, so is using the old English system of units That means that your ##g## is wrong; you need to use the value in English units also, or else convert the block's mass to kg.

There is also the issue of whether you should multiply by ##g## at all. In the USA at least, a 50 lb block exerts a force of 50 lb in Earth's sea-level gravity, so the ##g## is already incorporated in the 50 lb figure. (Basically, it is describing a mass by its force and mixing up the two concepts, but in a standard and understood way.)
 
Last edited:
  • Like
Likes   Reactions: Unicow, SammyS and BvU
Ray Vickson said:
Whoa! The question said the block weighs 50 lb, so is using the old English system of units That means that your ##g## is wrong; you need to use the value in English units also, or else convert the block's mass to kg.

There is also the issue of whether you should multiply by ##g## at all. In the USA at least, a 50 lb block exerts a force of 50 lb in Earth's sea-level gravity, so the ##g## is already incorporated in the 50 lb figure. (Basically, it is describing a mass by its force and mixing up the two concepts, but in a standard and understood way.)

Yeah, I ended up figuring that out sooner or later haha. Stupid mistake leading to almost an hour of checking over my work mostly. The answer was so simple. Thank you though haha. I tried to incorporate what we used in class but failed to recognize that we used kg in class vs lb.
 

Similar threads

  • · Replies 35 ·
2
Replies
35
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 2 ·
Replies
2
Views
14K
  • · Replies 6 ·
Replies
6
Views
13K
Replies
1
Views
2K
Replies
6
Views
3K