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

In summary: So I went back and double checked.In summary, a 50 lb block exerts a force of 50 lb in Earth's sea-level gravity, so the g is already included in the 50 lb figure.
  • #1
Unicow
14
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
 
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  • #2
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 !
 
  • #3
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:
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  • #4
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.
 

1. What is the formula for calculating the force on two cables with weight attached using theta?

The formula for calculating the force on two cables with weight attached using theta is F = 2Wsin(θ), where F is the force, W is the weight, and θ is the angle between the two cables.

2. How do I determine the angle theta in this scenario?

The angle theta can be determined by using trigonometric functions such as sine, cosine, or tangent. You can also use a protractor to directly measure the angle between the two cables.

3. What is the significance of the force on the cables in this situation?

The force on the cables is important because it determines the tension in the cables. The tension is crucial in keeping the weight attached to the cables in place and preventing any potential accidents.

4. Can the force on the cables be greater than the weight attached?

Yes, it is possible for the force on the cables to be greater than the weight attached. This can happen if the angle theta is less than 90 degrees, as the sine function can produce values greater than 1.

5. How does the addition of a weight affect the force on the cables?

The addition of a weight will increase the force on the cables. This is because the weight itself is exerting a downward force, which must be countered by an equal and opposite force from the cables. This results in an increase in tension and therefore an increase in the force on the cables.

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