Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 1 (Q.No. 6)
6.
The velocity ratio in case of an inclined plane inclined at angle θ to the horizontal and weight being pulled up the inclined plane by vertical effort is
Discussion:
58 comments Page 6 of 6.
Rajesh said:
6 years ago
What is the velocity ratio?
Pruthvi said:
6 years ago
I too agree, the correct answer is option D.
Ashish kukade said:
5 years ago
If weight is resolved w sin θ acts parallel to plane.
Then,
P= w sinθ.
MA = W/P = 1/ sin θ = Cosecθ.
Then,
P= w sinθ.
MA = W/P = 1/ sin θ = Cosecθ.
(3)
Jishnu said:
5 years ago
If x is the distance moved by the load on the inclined plane then the vertical distance moved by effort is also x. Now the velocity ratio is the distance moved by the effort to the distance moved by the load in the direction of effort.
That is x/x sin a = cosec a.
That is x/x sin a = cosec a.
(1)
Sangram mandal said:
4 years ago
A body is on an inclined plane means hypotension of right-angle triangle and effort applied upward means perpendicular of the triangle, velocity ratio is the distance by effort/distance by weight,
VR=P/h=h * sinθ/h = sinθ.
VR=P/h=h * sinθ/h = sinθ.
(4)
Abhishek said:
4 years ago
Anyone explain it in simple way to get it.
(6)
B.Lavanya said:
4 years ago
Here velocity ratio inclined plane at θ/2 to horizontal.
Here sinθ/2=hypothesis.
Here sinθ/2=hypothesis.
(6)
Md Irfan said:
2 years ago
The plate is inclined to horizontal plane. Therefore the horizontal component of force will be Cosθ whereas the vertical component will become Sinθ.
(15)
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