Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion
Discussion Forum : Hydraulics and Fluid Mechanics - Section 2 (Q.No. 26)
26.
An open tank containing liquid is made to move from rest with a uniform acceleration. The angle 0 which the free surface of liquid makes with the horizontal is such that (where a = Horizontal acceleration of the tank, and g = Acceleration due to gravity)
Discussion:
3 comments Page 1 of 1.
Muhammad Waqas said:
9 years ago
Consider XY plane.
Assume the motion is in +X direction it means acceleration is in +X direction.
The gravity acts in -Y direction.
Now make a vector diagram;
The required angle is in between base (x-axis) and hypotenuse. Right?
Now,
As we know that tan(x) = perpendicular/base.
So
Tan(x) = g/a.
Assume the motion is in +X direction it means acceleration is in +X direction.
The gravity acts in -Y direction.
Now make a vector diagram;
The required angle is in between base (x-axis) and hypotenuse. Right?
Now,
As we know that tan(x) = perpendicular/base.
So
Tan(x) = g/a.
Jay said:
9 years ago
Here angle is to be measured with the free surface, so answer will be g/a.
Amit said:
1 decade ago
Assuming motion in XZ plane:
tan(@) = (Acceleration in X axis)/((Gravity+ acceleration in z axis).
Here, acceleration in z axis = 0.
tan(@) = (Acceleration in X axis)/((Gravity+ acceleration in z axis).
Here, acceleration in z axis = 0.
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