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 1 of 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)

Suresh Srinivasalu said:   7 years ago
Velocity ratio=height of inclined plane/length of the inclined plane,
= h/l,
= sin(θ).
(8)

B.Lavanya said:   4 years ago
Here velocity ratio inclined plane at θ/2 to horizontal.
Here sinθ/2=hypothesis.
(6)

Abhishek said:   4 years ago
Anyone explain it in simple way to get it.
(6)

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.
(4)

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.
(3)

Venkata sai kumar said:   7 years ago
The inclined weight should transform to the vertical and horizontal axis.
The inclined plane makes θ with the horizontal axis.
Then weight on x axis= wcosθ.
Weighht on y axis= wsinθ.
Velocity ratio= wsinθ/wcosθ.
= Tanθ.
(1)

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.
(1)

Omkar said:   9 years ago
I think tan(theta) is correct.

Varun said:   9 years ago
But how? Please explain in detail.


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