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 4 of 6.
Sivaprakasam said:
9 years ago
I agree that the answer is sinθ.
Varun said:
9 years ago
But how? Please explain in detail.
S.k.gupta said:
9 years ago
sinθ is right answer. I agree.
Vinayak karma s said:
9 years ago
Velocity ratio = l/h, so its 1/sinθ -> cosecθ.
I think its cosecθ.
I think its cosecθ.
Kedar said:
9 years ago
I think tanθ is correct.
Aravindhan5223@gmail.com said:
9 years ago
Thank you all for explaining the answer.
Sathish said:
8 years ago
Thank you for the given explanation.
Uday said:
8 years ago
Inclined plane/upwards weight sinθ.
Horizontal= cosθ.
Vertical = sinθ.
Horizontal= cosθ.
Vertical = sinθ.
Sravan said:
8 years ago
Velocity ratio (Vr) = displacement by effort / displacement by force.
While pulling a block on an inclined plane, there is a vertical effort made by man to pull it through the pulley (which is a vertical plane, opposite to angle) By this force the block moves some displacement (which is on an inclined plane).
While pulling a block on an inclined plane, there is a vertical effort made by man to pull it through the pulley (which is a vertical plane, opposite to angle) By this force the block moves some displacement (which is on an inclined plane).
Yashwant said:
8 years ago
Suppose distance moved by a load on an inclined plane is x meter. So, the distance moved by effort is x sin θ. Now velocity ratio is equal to x sin θ upon x so hear sin θ is the right answer.
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