A uniform solid hemisphere of weight
W , radius
r and centre of mass distant
3r/8 from the centre of its plane face is placed with its circular face in contact with a smooth plane inclined at
ω to the horizontal. Equilibrium is maintained by a force
P tangential to the curved surface and in the vertical plane containing the centre of mass and line of greatest
slope of the plane. The direction of
P makes an
angle β with the vertical.
- Find P in terms of W, ω and β
- If the magnitude of the reaction of the plane on the hemisphere is R, show that:
R =
W sin
β cosec (
ω + β)
- Find the distance of the line of action of R from the centre of the circular face of the hemisphere
I have copied a link illustrating how the system of forces look-hope you can read/understand it!. Can anyone have a look and let me know if this can be solved by geometry, taking moments and give a clue etc. Please note the Reaction line is at right angles to the inclined plane because it is a smooth plane, please can you label this as C, which I have missed of the diagram. The three forces meet above the hemisphere where equilibrium is maintained by force P. The question states the force acts in the vertical plane and line of greatest slope. This means if you can imagine looking at the plane from the front projection the force will be acting vertically and tangentially to the centre of the hemisphere ie line of greatest slope and not and some skewed angle. Geometrically I struggle and I'm sure someone will be able to find out angles/lengths in order to solve it.
This is an interesting problem. The (3/8)r center of mass is what one would expect for a hemisphere of uniform density. The value of angle
β is not just a function of the slope of the plane, angle
ω, but also of the density of the object. The density of the object obviously affects its radius. As the object becomes more dense (assume a fixed mass), r shrinks, and
β shrinks along with it.
Let’s explore how this affects the relative magnitudes of the forces acting on the object. The weight of the object, W, will remain constant and always act vertically. Let’s draw a force diagram a bit differently than you did on your attachment. Start by drawing your plane and hemisphere as before. Draw W pointing downward from the center of mass (as you did before), then add a dotted-line vector, W’, of equal magnitude and opposite direction (upward). The tip of this upward vector defines a distinct point above the hemisphere. It is not at an arbitrary height as you have drawn in your original diagram. From this point above the hemisphere, draw a tangent line to the circle. Next, draw a line perpendicular to the plane and passing through the center of mass. Extend that line so that it intersects the tangent line.
Now we draw our two vectors P and R. For R, start the vector at the center of mass and end it at the tangent line. For P, start at the tip of R and extend it to the point above the hemisphere.
Your W, P, and R vectors are now drawn proportionately, and all forces cancel out for a net force of zero. The triangle they form has an angle of
β at the top and an angle of
ω at the bottom. This triangle should aid in answering the first two problem questions.
General commentary: As you can see from studying this diagram, the relative sizes of P and R will change as the radius of the hemisphere changes. As mentioned above, the radius is a function of density, so one might ask, why does the density not appear in the problem. The answer is that this is taken care of by the angle
β. Changes in
β will account for changes in density.
To tackle problem three, finding the distance, d, of R from the center of the sphere, I suggest summing the moments about the center of the sphere in which the hemisphere exists, i.e., the point on the plane that is (3/8)r from the center of mass. P is always r distance from this point, so its moment contribution is Pr in the clockwise (CW) direction. The weight will act in the CCW direction with moment of (W)((3/8)r)(sin
ω). The moment contribution of the plane reaction force, R, will be simply Rd. We know R from problem two:
R =
W sin
β cosec (
ω + β). The sum of these three moments must be zero. Just write the equation and solve for d.
Hope that helps.
