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 2 of 6.
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
Rounak said:
1 decade ago
@Sourav Kundu.
You can be right but in this question angle is made b/w horizontal and inclination.
So cos(90°-x) = sin x.
So right answer is (A).
You can be right but in this question angle is made b/w horizontal and inclination.
So cos(90°-x) = sin x.
So right answer is (A).
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)
Saifullah Khan Afridi said:
8 years ago
distance moved by effort= hyp.sinθ
distance moved by load= hyp
V.R= distance moved by effort/distance moved by load
V.R= sinθ.
distance moved by load= hyp
V.R= distance moved by effort/distance moved by load
V.R= sinθ.
Suresh Srinivasalu said:
7 years ago
Velocity ratio=height of inclined plane/length of the inclined plane,
= h/l,
= sin(θ).
= h/l,
= sin(θ).
(8)
Shiva rao said:
1 decade ago
Opposite side= sinθ.
Base= cosθ.
Because opp.side/hypo=sinθ(hypo=1).
sin2θ+cos2θ=1.
Base= cosθ.
Because opp.side/hypo=sinθ(hypo=1).
sin2θ+cos2θ=1.
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).
Haryhor. said:
1 decade ago
@Bibin S.
Velocity Ratio can be defined as the ratio of distance moved by effort to the distance moved by load.
Velocity Ratio can be defined as the ratio of distance moved by effort to the distance moved by load.
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.
Then,
P= w sinθ.
MA = W/P = 1/ sin θ = Cosecθ.
(3)
Unkown said:
10 years ago
It is simply inclination with respect to X & Y axis. So angle made with Y axis is Sin θ.
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