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.
Sikiru Abdullahi Ademola said:
1 decade ago
Option D is the correct answer.
V.R = 1/sin theta = cosec theta.
V.R = 1/sin theta = cosec theta.
Anurakti said:
1 decade ago
Didn't get any of the explanation. Can anyone explain it in a simpler way?
Sagar bodar said:
1 decade ago
Please state the condition regarding friction.
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).
Satish bhalke said:
1 decade ago
I think it should cosθ because weight is acting is down.
Ravi.b said:
1 decade ago
I think option C is correct answer because he is asking the velocity ratio. So we should resolve the given force in to two components x-component is mgcosθ, y-components is mgsinθ. So the ratio is tanθ.
Jegan said:
1 decade ago
What is the difference between effort and load?
Hemendra said:
1 decade ago
Can anyone explain briefly?
Karthik said:
10 years ago
Weight acts in vertical direction. So, sin 0.
Touqueer said:
10 years ago
Please answer me this question with the help of diagram.
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