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A thin plane lamina, of area A and weight W, slides down a fixed plane inclined to the vertical at an angle a and maintains a uniform gap & from the surface of the plane, the gap being filled with oil of constant viscosity u. The terminal velocity of the plane is?
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A thin plane lamina, of area A and weight W, slides down a fixed plane...
Let's start by drawing a diagram of the situation:
We have a fixed plane inclined at an angle a to the vertical, and a thin plane lamina of weight W and area A sliding down it with a uniform gap between the lamina and the plane.
Now, let's resolve the weight of the lamina into two components: one perpendicular to the plane and one parallel to it. The perpendicular component is Wcos(a), and the parallel component is Wsin(a).
The perpendicular component of the weight is balanced by the normal reaction force of the plane, N. Therefore, we have:
N = Wcos(a)
The parallel component of the weight causes the lamina to slide down the plane. The force of friction between the lamina and the plane, F, acts in the opposite direction to this motion. Therefore, we have:
F = Wsin(a)
Now, let's consider the forces acting on the lamina in the direction perpendicular to the plane. We have:
N - Mg = 0
where M is the mass of the lamina and g is the acceleration due to gravity. Therefore, we have:
Mg = N = Wcos(a)
Now, let's consider the forces acting on the lamina in the direction parallel to the plane. We have:
F = Ma
where a is the acceleration of the lamina down the plane. Therefore, we have:
Wsin(a) = Ma
Solving for a, we get:
a = (Wsin(a))/M
Finally, let's use the uniform gap between the lamina and the plane to find an expression for the angle of inclination of the plane. We have:
tan(a) = gap/length of plane
where the length of the plane is L. Therefore, we have:
a = tan^-1(gap/L)
Substituting this expression for a into our expression for a, we get:
a = (Wsin(tan^-1(gap/L)))/M
This is the final expression for the acceleration of the lamina down the plane in terms of its weight, area, mass, length, and the uniform gap between the lamina and the plane.
We have a fixed plane inclined at an angle a to the vertical, and a thin plane lamina of weight W and area A sliding down it with a uniform gap between the lamina and the plane.
Now, let's resolve the weight of the lamina into two components: one perpendicular to the plane and one parallel to it. The perpendicular component is Wcos(a), and the parallel component is Wsin(a).
The perpendicular component of the weight is balanced by the normal reaction force of the plane, N. Therefore, we have:
N = Wcos(a)
The parallel component of the weight causes the lamina to slide down the plane. The force of friction between the lamina and the plane, F, acts in the opposite direction to this motion. Therefore, we have:
F = Wsin(a)
Now, let's consider the forces acting on the lamina in the direction perpendicular to the plane. We have:
N - Mg = 0
where M is the mass of the lamina and g is the acceleration due to gravity. Therefore, we have:
Mg = N = Wcos(a)
Now, let's consider the forces acting on the lamina in the direction parallel to the plane. We have:
F = Ma
where a is the acceleration of the lamina down the plane. Therefore, we have:
Wsin(a) = Ma
Solving for a, we get:
a = (Wsin(a))/M
Finally, let's use the uniform gap between the lamina and the plane to find an expression for the angle of inclination of the plane. We have:
tan(a) = gap/length of plane
where the length of the plane is L. Therefore, we have:
a = tan^-1(gap/L)
Substituting this expression for a into our expression for a, we get:
a = (Wsin(tan^-1(gap/L)))/M
This is the final expression for the acceleration of the lamina down the plane in terms of its weight, area, mass, length, and the uniform gap between the lamina and the plane.
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A thin plane lamina, of area A and weight W, slides down a fixed plane inclined to the vertical at an angle a and maintains a uniform gap & from the surface of the plane, the gap being filled with oil of constant viscosity u. The terminal velocity of the plane is?
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A thin plane lamina, of area A and weight W, slides down a fixed plane inclined to the vertical at an angle a and maintains a uniform gap & from the surface of the plane, the gap being filled with oil of constant viscosity u. The terminal velocity of the plane is? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A thin plane lamina, of area A and weight W, slides down a fixed plane inclined to the vertical at an angle a and maintains a uniform gap & from the surface of the plane, the gap being filled with oil of constant viscosity u. The terminal velocity of the plane is? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thin plane lamina, of area A and weight W, slides down a fixed plane inclined to the vertical at an angle a and maintains a uniform gap & from the surface of the plane, the gap being filled with oil of constant viscosity u. The terminal velocity of the plane is?.
A thin plane lamina, of area A and weight W, slides down a fixed plane inclined to the vertical at an angle a and maintains a uniform gap & from the surface of the plane, the gap being filled with oil of constant viscosity u. The terminal velocity of the plane is? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A thin plane lamina, of area A and weight W, slides down a fixed plane inclined to the vertical at an angle a and maintains a uniform gap & from the surface of the plane, the gap being filled with oil of constant viscosity u. The terminal velocity of the plane is? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thin plane lamina, of area A and weight W, slides down a fixed plane inclined to the vertical at an angle a and maintains a uniform gap & from the surface of the plane, the gap being filled with oil of constant viscosity u. The terminal velocity of the plane is?.
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