A Symmetrical lamina has the dimension shown Prove that the

A Symmetrical lamina has the dimension shown. Prove that the distance from the edge b to the center of mass in: (2a + b)L/3(a + b) (b) Check the answer above by showing that is reduces to the correct result is limiting cases (e.g. when a = 0 and when a = b) 371 A uniform smooth plank of weight a and length 2a is hinged to the bottom horizontal edge of a smooth, fixed plane inclined at angle beta to the horizontal. A sphere of radius b/a and weight 2w is placed between the plank and the plane. Assume no friction. Prove that, in the position of equilibrium, the angle theta between the plank and the plane will be given by the equations: 2 sin^2theta/2cos(beta + beta)=sinbeta 372 A hemispherical bowl is fixed with its rim in a horizontal plane, Into the bowl is laid a rod of lenght equal to the diameter of the bowl. The rod is uniform. There is no friction at either point of contact. Find the angle between the rod and the horizontal when the rod assumes its equilibrium position.

Solution

lET THE NORMAL FORCE ACT ON THE PLANK AT A DISTANCE X FROM THE ORIGiN

x*N = a*W*cos(theta + beta) -- (1)
Also, from geometry tan(theta/2) = a/2x
x = acot(theta/2) /2 -- (2)
plug in 1
N = 2Wcos(theta + beta)tan(theta/2)

Now, Force balance on the ball:

Ncos(theta + beta) + 2W - N\'cos(beta) = 0
and Nsin(theta+beta) = N\'sin(beta)

or

N\' =[ 2Wcos(theta + beta)tan(theta/2)cos(theta + beta) + 2W ]/cos(beta)
or [ 2Wcos(theta + beta)tan(theta/2)cos(theta + beta) + 2W ]/cos(beta) = [2Wcos(theta + beta)tan(theta/2) sin(theta + beta)]/sin(beta)
or [ cos(theta + beta)tan(theta/2)cos(theta + beta) + 1 ]*sin(beta) = [cos(theta + beta)tan(theta/2) sin(theta + beta)]*cos(beta)
or cos^2(theta+beta)sin(theta/2)sin(beta) + sin(beta)*cos(theta/2) = sin(theta+beta)cos(theta + beta)sin(theta/2)cos(beta)
or sin(beta)*cos(theta/2) = sin(theta+beta)cos(theta + beta)sin(theta/2)cos(beta) - cos^2(theta+beta)sin(theta/2)sin(beta)
or sin(beta)*cos(theta/2) = sin(theta/2)*cos(theta+beta)*[sin(theta+beta)cos(beta) - cos(theta+beta)sin(beta)]
or sin(beta)*cos(theta/2) = sin(theta/2)*cos(theta+beta)*[sin(theta)]
or sin(beta) = 2sin(theta/2)*cos(theta+beta)*[sin(theta/2)]