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.
Uday said:
8 years ago
Inclined plane/upwards weight sinθ.
Horizontal= cosθ.
Vertical = sinθ.
Horizontal= cosθ.
Vertical = sinθ.
Sathish said:
8 years ago
Thank you for the given explanation.
Aravindhan5223@gmail.com said:
9 years ago
Thank you all for explaining the answer.
Kedar said:
9 years ago
I think tanθ is correct.
Vinayak karma s said:
9 years ago
Velocity ratio = l/h, so its 1/sinθ -> cosecθ.
I think its cosecθ.
I think its cosecθ.
S.k.gupta said:
9 years ago
sinθ is right answer. I agree.
Varun said:
9 years ago
But how? Please explain in detail.
Sivaprakasam said:
9 years ago
I agree that the answer is sinθ.
Omkar said:
9 years ago
I think tan(theta) is correct.
Geek said:
9 years ago
The answer should be cosec (theta).
The distance moved by effort is the length of the inclined plane, the distance moved by the load is the vertical distance if you resolve these you get X/Xsin (theta) which is 1/sinX.
The distance moved by effort is the length of the inclined plane, the distance moved by the load is the vertical distance if you resolve these you get X/Xsin (theta) which is 1/sinX.
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