Mechanical Engineering - Engineering Mechanics - Discussion

As effort has been applied vertically , hence displacement is calculated along the direction of force. Considering the mass does not get detached from the inclined plane, the mass is constrained to move along the inclined plane. Therefore displacement along the hypotenuse would be 1 unit( considering a triangle of 1 unit hypotenuse). Consequently, displacement along the effort would be sin(theta). VR= displacement by effort/ displacement by load, = sin(theta).

Its the same as screw thread principle.

Velocity ratio (Vr) = displacement by effort / displacement by force.

While pulling a block on an inclined plane, there is a vertical effort made by man to pull it through the pulley (which is a vertical plane, opposite to angle) By this force the block moves some displacement (which is on an inclined plane).

The inclined weight should transform to the vertical and horizontal axis.
The inclined plane makes θ with the horizontal axis.
Then weight on x axis= wcosθ.
Weighht on y axis= wsinθ.
Velocity ratio= wsinθ/wcosθ.
= Tanθ.

If x is the distance moved by the load on the inclined plane then the vertical distance moved by effort is also x. Now the velocity ratio is the distance moved by the effort to the distance moved by the load in the direction of effort.

That is x/x sin a = cosec a.

A body is on an inclined plane means hypotension of right-angle triangle and effort applied upward means perpendicular of the triangle, velocity ratio is the distance by effort/distance by weight,

VR=P/h=h * sinθ/h = sinθ.

Velocity ratio is the ratio of distance moved by effort to distance moved by force, in case a weight is pulled using a pulley the vertical side is the distance moved by effort, the inclined side is the distance moved by the force.

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.

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.

Due to the effort the block moves up the inclined plane.
Due to the effort the velocity created along the plane is v.
Velocity created due to vertical effort is v siny.
Therefore VR {vertical effort to effort} = siny.

Suppose distance moved by a load on an inclined plane is x meter. So, the distance moved by effort is x sin θ. Now velocity ratio is equal to x sin θ upon x so hear sin θ is the right answer.