5 Three Boxes are located on three inclined surfaces with th

5. Three Boxes are located on three inclined surfaces with the friction. The friction coefficients are identical. Angles of inclined surfaces with the horizontal direction 0, 0,, and 0, are different. On the inclined surface with the angle the acceleration of the box is an. On the incline surface with the angle O2, the acceleration is az. The motion on the third surface happens without any acceleration. Find an expression for third angle in terms of az

Solution

Here,

theta1 , theta2 , theta3

for the angle theta1

a1 = g * sin(theta1) - u * g * cos(theta1) ----(1)

for the angle theta2

a2 = g * sin(theta2) - u * g * cos(theta2)

for the angle theta3

0 = g * sin(theta3) - u * g * cos(theta3)

tan(theta3) = u

from 1

u = (g * sin(theta1) - a1 )/(g * cos(theta1))

putting in the equation

tan(theta3) = (g * sin(theta1) - a1 )/(g * cos(theta1))

theta3 = arctan((g * sin(theta1) - a1 )/(g * cos(theta1)))

the expression for the third angle is arctan((g * sin(theta1) - a1 )/(g * cos(theta1)))