Question

A rope of negligible mass is stretched horizontally between two supports that are 2.81 m apart. When an object of weight 4190 N is hung at the center of the rope, the rope is observed to sag by 39.6 cm. What is the tension in the rope?

Answer 1

By applying parallelogram law of forces the tension on each half of the rope can be calculated.

........A........... .........C.......... ..........B.......


............... ............. ..D........ ....................


.............. .................W.. ........ ...........
AB is the length of the rope= 2.81 m
CD is the sag..... ................= 0.396 m
AD & DB are the semi-sections of the rope= 1.405 m
W is the weight suspended from the point D
This arrangement is the same as a set of three concurrent forces in equillibrium.
DW is the weight = 4190 N
DA & DB are the other two forces, the resultant f which will be equal to the weight suspended .
From the data given, the angle CDB can be found out to be eqal to 78°. So the angle ADB will be equal to 156°
As C is the central point, AD & DB are equal
From the formula for the resultant R in the parallelogram law of forces,
R = √(P²+Q²+2PQCos α)
As the forces DA & DB are equal, P will be equal to Q. Hence, the formula can be rewritten as:
R = √(2P²+2P²Cosα)
R = √{2P²(1+Cosα)}
R² = 2P²(1+Cosα)
P² = R² / {2(1+Cosα)}
P = √ [ R² / {2(1+Cosα)}]

P = √ [4190² / {2(1+Cos156°)}]
P = √[17556100 / {2(1- 0.9135)}]
P = √[17556100 / (2x.0865)]
P = √[17556100 / 0.173]
P= √70002312
P = 10073 N
As P & Q are equal, Q = 10073 N

Tension on DA segment = 10073 N
Tension on DB segment = 10073 N
Tension on DW segment = 4190 N