Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 3 (Q.No. 50)
50.
A rod 5 m in length is moving in a vertical plane. When it is inclined at 60° to horizontal, its lower end is moving horizontally at 3 m/sec and upper end is moving in vertical direction. The velocity of its upper end, is
Discussion:
2 comments Page 1 of 1.
Ihsan said:
5 years ago
The rod is moving in a way that it makes 60 angles with horizontal. It's rare, the end is moving horizontally with sqr 3 velocity which is the horizontal component of the resultant velocity. And the moving of the top end upward is a vertical component of the resultant velocity.
So, Let the resultant velocity is V, horizontal is Vh and Vertical is Vv and angle is ø,.
So, R cos ø = Vh,
R cos 60= sqr 3,
R= 3.46.
And R sin ø =Vv.
3.46 sin 60 = Vv
Vv= 3 m/s.
So, Let the resultant velocity is V, horizontal is Vh and Vertical is Vv and angle is ø,.
So, R cos ø = Vh,
R cos 60= sqr 3,
R= 3.46.
And R sin ø =Vv.
3.46 sin 60 = Vv
Vv= 3 m/s.
Sachib said:
8 years ago
It should be 3m/s.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers