The shape of a suspended cable for a uniformly distributed load over it is
A. Circular
B. Parabolic
C. Catenary
D. Cubic parabola
Answer: Option B
Solution (By Examveda Team)
The correct answer is B: Parabolic.Here's why:
* Uniformly Distributed Load: This means the load is spread evenly across the horizontal span of the cable (like the weight of a bridge deck hanging from the cable).
* Parabolic Shape: When a cable is subjected to a uniformly distributed load horizontally, the cable takes the shape of a parabola. This shape ensures that the tension in the cable is efficiently distributed to support the load.
* Catenary Shape (Why it's not the answer): A catenary curve is the shape a cable takes when it's only supporting its own weight (or a load distributed along the length of the cable itself, not horizontally). Think of a cable hanging freely between two points.
* Other Options: Circular and Cubic Parabola shapes are not naturally formed by suspended cables under common loading conditions like a uniform distributed load.
This Question Belongs to Civil Engineering >> Applied Mechanics And Graphic Statics
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Comments (1)
In case of S.H.M. the period of oscillation (T), is given by
A. $${\text{T}} = \frac{{2\omega }}{{{\pi ^2}}}$$
B. $${\text{T}} = \frac{{2\pi }}{\omega }$$
C. $${\text{T}} = \frac{2}{\omega }$$
D. $${\text{T}} = \frac{\pi }{{2\omega }}$$
The angular speed of a car taking a circular turn of radius 100 m at 36 km/hr will be
A. 0.1 rad/sec
B. 1 rad/sec
C. 10 rad/sec
D. 100 rad/sec
A body is said to move with Simple Harmonic Motion if its acceleration, is
A. Always directed away from the centre, the point of reference
B. Proportional to the square of the distance from the point of reference
C. Proportional to the distance from the point of reference and directed towards it
D. Inversely proportion to the distance from the point of reference
The resultant of two forces P and Q acting at an angle $$\theta $$, is
A. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{P}}\sin \theta $$
B. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta $$
C. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\tan \theta $$
D. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta } $$
E. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\sin \theta } $$
Correct option is c please correct it in your system