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
sin θ
cos θ
tan θ
cosec θ
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
58 comments Page 3 of 6.

Ravi said:   8 years ago
What is velocity ratio here?

A.G.Kesavasekaran said:   8 years ago
Due to the effort the block moves up the inclined plane.
Due to the effort the velocity created along the plane is v.
Velocity created due to vertical effort is v siny.
Therefore VR {vertical effort to effort} = siny.

Ramesh said:   8 years ago
A is the right answer. I agree.

Saifullah Khan Afridi said:   8 years ago
distance moved by effort= hyp.sinθ
distance moved by load= hyp
V.R= distance moved by effort/distance moved by load
V.R= sinθ.

SAta said:   8 years ago
Velocity ratio of an inclined plane= cosecθ.

Option D is correct.

Prashant said:   8 years ago
Option B is the correct answer.

Akash said:   7 years ago
It is Displacement/distance.

Rajesh said:   6 years ago
What is the velocity ratio?

Pruthvi said:   6 years ago
I too agree, the correct answer is option D.

Ravi.b said:   1 decade ago
I think option C is correct answer because he is asking the velocity ratio. So we should resolve the given force in to two components x-component is mgcosθ, y-components is mgsinθ. So the ratio is tanθ.


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