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 3 of 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)
Uday said:
8 years ago
Inclined plane/upwards weight sinθ.
Horizontal= cosθ.
Vertical = sinθ.
Horizontal= cosθ.
Vertical = sinθ.
Vinayak karma s said:
9 years ago
Velocity ratio = l/h, so its 1/sinθ -> cosecθ.
I think its cosecθ.
I think its cosecθ.
Jayant said:
9 years ago
When the angle is given with respect to horizontal, then in Y axis sin theta is taken.
Sourav Kundu said:
1 decade ago
I think option (B) because W and are make angle and weight is shown by cos.
Anurakti said:
1 decade ago
Didn't get any of the explanation. Can anyone explain it in a simpler way?
SAta said:
8 years ago
Velocity ratio of an inclined plane= cosecθ.
Option D is correct.
Option D is correct.
Yashavant said:
10 years ago
@Manoj.
Angle touch to horizontal axis i.e. base hence it is cos.
Angle touch to horizontal axis i.e. base hence it is cos.
Sikiru Abdullahi Ademola said:
1 decade ago
Option D is the correct answer.
V.R = 1/sin theta = cosec theta.
V.R = 1/sin theta = cosec theta.
Satish bhalke said:
1 decade ago
I think it should cosθ because weight is acting is down.
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