Question

Each of the following planar objects is placed, as shown in Fig. 2.13 (JUST ONE), between two frictionless circles of radius R. The mass density per unit area of each object is σ, and the radii to the points of contact make an angle θ with the horizontal. For each case, find the horizontal force that must be applied to the circles to keep them together. For what θ is this force maximum or minimum? An isosceles triangle with common side length L.

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Answer 1

From the given diagram, we can see
angle made by radii with point of contact and the horizontal is theta
circles are of radii R, mass densoty per unit area of each object being sigma
The triangle is isosceles with common side L

Now, since there is no friction, there must be contact forces acting
so, from symmetry, let the normal force between the triangle and the circle be B, and normal force between the two circles be C
then
From force balance on the circle
F = C + B*cos(theta)

From force balance on the triangle and applying force on gravity on it as well we get
also, inside angle of the isosceles triangle will be 2*theta
so area of triangle is 0.5*(2b)*h
here,
L*cos(theta) = h
L*sin(theta) = b
A = L^2*sin(theta)cos(theta)
2B*sin(theta) = sigma*A*g = sigma*L^2*sin(theta)cos(theta)*g
where g is acceleration due to gravity

hence
B = sigma*L^2*cos(theta)*g/2

also,
assuming C = 0 (the circles are just touching)
F = B*cos(theta) = sigma*L^2*cos^2(theta)*g/2

hence in case of the isosceles triangle of length L of the equal side, force F = sigma*L^2*cos^2(theta)*g/2 has to be applied to keep the mass stable
sigma is mass density per unit area, L is side of the triangle, theta ishte angle of contact points with horizontal , g is acceleration due to gravity

now, from the formula for F = sigma*L^2*cos^2(theta)*g/2
for maxima or minima, cos^2(theta) = 1, 0
so minimum F is required for theta = 90 deg ( when the triangle becomes a horizontal rod)
maximum required force is for theta = 0 deg (when the triangle becomes a vertical rod)