Question

7. The configuration below shows four identical objects of mass m hung from string of total length L. The points of support are evenly spaced a distance l apart as shown below and the angles L1 and L2 are indicated.

Answer 1

a. Find the tension in each section of the string in terms of 1, m and g.
b. Find an expression for the angle 2 as a function of 1.
c. Show that the distance D between the endpoints of the string can be expressed as:

Answer 2

\[\theta \] - missing squares

Answer 3

Check the attached file which shows various forces acting on first two masses { 01=theta1, 02=theta2} . Now the tension in the first piece if the string(T1) is equal to the tension in the fifth piece of the string because the entire system is symmetric since all the masses are equal. Similarly tension in second piece (T2) is same as that in fourth piece. Now we apply newton's second law on first two masses.

FOR MASS A IN VERTICAL DIRECTION

T1*sin 01 = mg + T2*sin 02 ......................................... (1)

FOR MASS B IN HORIZONTAL DIRECTION

T1*cos 01 = T2*cos 02 .............................................(2)

FOR MASS B IN VERTICAL DIRECTION

T2*sin 02 = mg ...........................................(3)

FOR MASS B IN HORIZONTAL DIRECTION

T2*cos 02 = T3 .............................................(4)


(a)
T1 is obtained by adding (1) and (3)
\[T _{1}= 2mg/\sin \theta1\]

T3 is obtained by adding (2) and (4) and then using the value of T1 obtained above
\[T _{3}= 2mg*\cot \theta1\]

T2 is obtained by squaring and adding (3) and (4) and then using T3 obtained above
\[T _{3}= mg \sqrt{1+4 \cot ^{2}\theta1}\]

(b)

02 can be obtained by dividing (3) by (4) and then using the value of T3 obtained \[\tan \theta2 =(\tan \theta1)/2\]


(c)

In triangle ABC
BC= AB*cos 01 = l *cos 01
In triangle BDE
ED =BD*cos 02 = l*cos 02

Clearly from the figure
D=2*BC+2*ED+ l
=2*l*cos 01+2*l*cos 02 + l
= l [2 cos 01 + 2 cos (tan-1(tan 02)) +1]
=L/5 [2 cos 01 +2 cos (tan-1(tan 02))+1]

which was to be proved.