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 5 of 6.
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.
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.
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.
Manoj said:
10 years ago
How you considered that the base as Cos theta?
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