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 6 of 6.
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.
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
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).
Unkown said:
10 years ago
It is simply inclination with respect to X & Y axis. So angle made with Y axis is Sin θ.
Arockia Ranjith said:
9 years ago
Which one is correct? Please explain the correct answer.
Jayant said:
9 years ago
When the angle is given with respect to horizontal, then in Y axis sin theta is taken.
Shara said:
9 years ago
Sin (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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