# Thread: Statics of a particle and tension of strings

1. ## Statics of a particle and tension of strings

Can someone help me with this problem?

A string is connected to two points A and D in a horizontal line and weights of 12 kg and
W kg are attached at points B and C. If AB, BC and CD make angles of 40◦, 20◦ and 50◦
respectively with the horizontal, calculate the tensions in the string and the weight W.

Can someone help me with this problem?

A string is connected to two points A and D in a horizontal line and weights of 12 kg and
W kg are attached at points B and C. If AB, BC and CD make angles of 40◦, 20◦ and 50◦
respectively with the horizontal, calculate the tensions in the string and the weight W.
Not only is it hard to imagine the figure, but it is also harder to solve the question if there are no given distances among or between the four points A,B,C,D.

The points B and C are hanging? The points A and D are the ends of the string and they are anchored to the horizontal "line"?

3. Yes i think the points B and C are hanging and i think the way to work it out is to rearrange the vectors into a triangle and use trigonometry to work out the sides but i can't do it. i agree its hard to figure it out without the distances. The answers given in the book are
32.97 kg wt, 26.88 kg wt, 39.29 kg wt,
W = 39.29 kg
if that helps to work it out

4. i think i've figured out the angles A=40◦, B=160◦, C=110◦, D=50◦ and the line AB would be 12kg wt? i don't know what to do from here though

5. actually no, AB wouldn't be 12kg wt, cos that would be the downwards force and they give three answers in the book for the three tensions of the strings. W would also be a vertically downwards force i think???

Can someone help me with this problem?

A string is connected to two points A and D in a horizontal line and weights of 12 kg and
W kg are attached at points B and C. If AB, BC and CD make angles of 40◦, 20◦ and 50◦
respectively with the horizontal, calculate the tensions in the string and the weight W.
Another look at your question here, another approach, and I found a way to solve it.

We have to use moments. I hope you know how moments work in Mechanics.
Moment is (force)*(perpendicular distance).
The perpendicular distance is the moment arm of the force.
From a point where you are taking moments, find the moment arm of the force by solving for the perpendicular distance from the point that is perpendicular to the line of action of the force.

Forces whose lines of action pass through a point, have no moment about that point.

The distances are not needed because the configuration of the system...the given angles.... will fix the points in places. Meaning, with the given angles, there is only one configuration that will satisfy those. Well, there are infinitely many but they are all similar.
So we let x = length of BC.

To solve for the W and the tensions on the 3 segments of the string, we find ways to take moments about some points where there are minimum of unknown forces. We find points where the lines of actions of most of the forces intersect.

If we extend AB and DC, they will intersect at point , say, P.
Since angle DAP = 40deg, and angle ADP is 50 deg, then the angle at P is a right angle.

Draw a vertical line through P.
Draw a horizontal line through B.
Draw a vertical line through C.
Call the points of intersections along the horizontal line through B as M and Q.
So the horizontal distance between the 12 kg and the W is BMQ.
The horizontal distance between the 12 kg and P is BM. Call that "a".
The horizontal distance between the W and P is MQ. Call that b.

Let BP = u, and CP = v

See if you can find the ways why
>>>angle CPM = 40 deg
>>>angle BPM = 50 deg
>>>angle PBC = 20 deg
>>>angle PCB = 70 deg

So, in right triangle BPC,
u = (BC)cos(angle PBC) = x*cos(20deg) = (0.93969)x
v = x*sin(angle PBC) = x*sin(20deg) = (0.34202)x

In right triangle BMP,
a = u*sin(angle BPM) = (0.93969 x)sin(50deg) = (0.71984)x

In right triangle BQC,
b = x*cos(20deg) -a = (0.21985)x

Okay, we are ready to take moments now.

Cut the string below points A and D to make a free-body diagram, FBD, of the system.

Let T = tension at string BA; S = tension at string BC; R = tension at string CD.

To find W, we take moments at point P....because the T and R will intersect there, while the S is a non-entity, a "sleeping beauty", according to the FBD.

As you know, in a system in equilibrium, the moment about any point is zero.

Summation of moments at point P is zero.
12kg *a = w*b
W = 12a/b = 12(0.71984 x) /(0.21985 x) = 39.29 kg ---------answer.

(I see that your book gave the answers in kg. Even for tensions. Okay, let's follow those. In reality, though, forces must be in newtons.)

To get T, we take moments at point C.
At point C, the W and R have no moments.
The moment arm of T from point C is v....remember that DP is perpendicular to AB.
The moment arm of the 12 kg is (a+b).
So,
T*v = 12*(a+b)
T = 12(0.71984x +0.21985x) /(0.34202x) = 32.97 kg ----------answer.

To find R, take moment at point B.
W*(a+b) = R*u
R = (39.29)(0.71984x +0.21985x) /(0.93969x) = 39.29 kg ---- answer.

To find S, there are no more points where we can eliminate many forces, so we will use the summation of vertical forces, or of horizontal forces, is zero.

Say, from the uncut system, we create another FBD. Cut the AB string and the BC string.
So our FBD now consist of the T, the 12kg, and the S.
Summation of vertical forces is zero,
T*sin(40deg) = 12 +S*sin(20deg)
S = [(32.97)sin(40deg) -12] /sin(20deg) = 26.88 kg --------answer.