# Thread: Help with Force Problem.

1. ## Help with Force Problem.

Two traffic signals are temporary suspended from a cable as shown. Knowing that the signal B weighs 300 Newtons, determine the weight of signal C.

(I know how to compute vectors and how to find them. All I need to know is which force vectors are used. And what is the equation. Just write out the vectors used.)

2. Originally Posted by ThePerfectHacker
Two traffic signals are temporary suspended from a cable as shown. Knowing that the signal B weighs 300 Newtons, determine the weight of signal C.

(I know how to compute vectors and how to find them. All I need to know is which force vectors are used. And what is the equation. Just write out the vectors used.)
I assume that all the strings are tight and the whole system is stationary i.e. the balls B and C are in equilibrium.
Let the tension in the part AB of the string be $T_{AB}$(I don't know how to make overhead arrow, just assume it a vector). Its direction is from A to B. $T_{BA}$ represents the same tension with direstion reversed.
Equations are
$T_{BA}+T_{BC}+W_B=0$
$T_{CB}+T_{CD}+W_C=0$
WEIGHTS ARE DIRECTED DOWNWARDS

Malay

3. Good that I understand.
But how do I solve them?
There are 5 variables over here and 2 equations.

4. Originally Posted by ThePerfectHacker
Good that I understand.
But how do I solve them?
There are 5 variables over here and 2 equations.
Write them out in full, you have two 2-vector equations, which
makes 4 scalar equations, and the unknowns are the absolute values of
the three tensions and the weight of the signal at C.

RonL

5. Originally Posted by ThePerfectHacker
Good that I understand.
But how do I solve them?
There are 5 variables over here and 2 equations.
Let us analyse first equation(you have two unknowns)
Draw a 2D coordinate system in the plane with B as origin and x-axis parallel to the basei.e. road I think. you know the angles of the three vectors with the axes. For b to be in equilibrium, components of the forces along both the axes is zero(you will get two linear equations for two unknowns).
After solving the first equation, you will two unknowns in the second.

Keep Smiling
Malay

6. Okay I got it now, thank you.
I wrote them out in components form, thus I get 4 equations and 4 variables.

7. ## Hooray for applied maths

Looks like we have a good thing going here with the applied math section.

8. Originally Posted by ThePerfectHacker
Two traffic signals are temporary suspended from a cable as shown. Knowing that the signal B weighs 300 Newtons, determine the weight of signal C.

(I know how to compute vectors and how to find them. All I need to know is which force vectors are used. And what is the equation. Just write out the vectors used.)
There are four unknown forces--the tensions on the 3 sections of the cable, and that one due to the weight of signal C.
To find that force due to the weight of signal C, find a way to eliminate or to not include the 3 tensile forces, leaving you to deal only with the given 300N signal B and the unknown signal C.
This said way is to take moments at the intersection of the tensile forces at AB and CD. The 3rd tension is not even a problem here.
The system is static, and in equilibrium. Imagine cutting the cable at two points: anywhere along AB and along CD, so you have a free body diagram of the system.
Extend AB and CD to intersect at point, say, E. The figure now looks like a triangle BEC. Draw the forces acting at points B and C.
At point B: the tensile force along AB, and the downward vertical force 300 newtons.
At point C: the tensile force along CD, and the downward vertical force, say, X newtons.

Since the FBD is in equilibrium, then the moments of all the forces about any point are equal to zero. (Moment = force times perpendicular distance.) Take the moments about point E.
---Moments due to the two tensile forces are zero--because the perpendicular distance of each from point E is zero.

---Moment due to the 300N at point B:
The direction of the 300N is vertical, so its perpendicular distance from point E is a horizontal distance. Call this horizontal distance between points B and E as "a" meters.
Hence, moment due to 300N is (300*a) newton-meters, counterclockwise.

---Moment due to the X newtons at point C:
The direction of the X is vertical also, so its perpendicular distance from point E is a horizontal distance too. Since the given horizontal distance between points B and C is 3.4 m, then the horizontal distance of X from point E is (3.4 -a) meters.
Hence, moment due to the X is X*(3.4 -a) newton-meters, clockwise.

Then, the summation of moments being zero, all clockwise moments equal all counterclokwise moments,
X(3.4 -a) = 300*a
So,
X = (300a)/(3.4 -a) newtons ------------(1)

You need to find "a" then. One way of doing that is by using "equations of the lines" of AB and CD, by their point-slope forms.
(y -y1) = m(x -x1) ----(i)
m = (y2 -y1)/(x2 -x1) = (y1 -y2)/(x1 -x2) ---(ii)
Blah, blah, blah.

You should get a = 0.8353 meter.

Therefore, depending on what precision you want, you should get X = about 92 to 98 newtons. ------------answer.

(All the given dimensions are in one decimal place only, so you are to use numbers in your computations in one decimal place only?)

9. Originally Posted by ThePerfectHacker
Good that I understand.
But how do I solve them?
There are 5 variables over here and 2 equations.
You can apply Lami's Theorem

Keep Smiling
Malay