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.
Shara said:
9 years ago
Sin (theta) is correct.
Jayant said:
9 years ago
When the angle is given with respect to horizontal, then in Y axis sin theta is taken.
Arockia Ranjith said:
9 years ago
Which one is correct? Please explain the correct answer.
Unkown said:
10 years ago
It is simply inclination with respect to X & Y axis. So angle made with Y axis is Sin θ.
Ranjeet Kumar said:
10 years ago
Since, v = u+at,
v1= gsin (theta)t, u = 0, a = g cos(90-theta) = g sin(theta).
v2 = gt.
So, v2/v1 = sin(theta).
v1= gsin (theta)t, u = 0, a = g cos(90-theta) = g sin(theta).
v2 = gt.
So, v2/v1 = sin(theta).
Amit kumar said:
10 years ago
V.R = (Distance moved by effort)/(Distance moved by lode).
Since effort is vertical so it will be P sinθ and load will be:
I = (p sinθ)^2+(p cos θ)^2
Since effort is vertical so it will be P sinθ and load will be:
I = (p sinθ)^2+(p cos θ)^2
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.
Manoj said:
10 years ago
How you considered that the base as Cos theta?
Touqueer said:
10 years ago
Please answer me this question with the help of diagram.
Karthik said:
10 years ago
Weight acts in vertical direction. So, sin 0.
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