Problem 8.159 C A homogeneous hemisphere of radius r is placed on an incline as shown. Knowing that the coefficient of static friction between the hemisphere and the incline is 0.30, determine (a) the value of b for which sliding impends, (b) the corresponding value of q. G q b

Solving Problems on Your Own A homogeneous hemisphere of radius r is placed on an incline as shown. Knowing that the coefficient of static friction between the hemisphere and the incline is 0.30, determine (a) the value of b for which sliding impends, (b) the corresponding value of q. C G q b When all the applied forces and the coefficient of friction are known, and you must determine when sliding occurs : a. Write the equations of equilibrium to determine N and F . b. Calculate the maximum friction force. For a two-force body it may be easier to use the resultant force R and the angle f it makes with the normal to the surface.

Problem 8.159 Solution Write the equations of equilibrium to determine N and F . Calculate the maximum friction force. For a two-force body it may be easier to use the resultant force R and the angle f it makes with the normal to the surface. W C G q b A R fs We have a two-force body. For sliding to impend R forms an angle fs with the incline. fs = tan-1 0.30 = 16.70o b = 16.70o 3 8 Geometry: GC = r (see fig. 5.21) AC = r

sin (180o - q) sin fs AC GC = sin (180o - q) = sin fs = sin 16.70o Problem 8.159 Solution fs = tan-1 0.30 = 16.70o b = 16.70o Write the equations of equilibrium to determine N and F . Calculate the maximum friction force. For a two-force body it may be easier to use the resultant force R and the angle f it makes with the normal to the surface. W C G q b A R fs ACG = q - fs C Triangle ACG: q AGC = 180o - q G fs sin (180o - q) AC sin fs GC fs Law of sines = AC GC r 3r/8 sin (180o - q) = sin fs = sin 16.70o A

sin (180o - q) = sin fs = sin 16.70o sin (180o - q) = 0.76626 Problem 8.159 Solution Write the equations of equilibrium to determine N and F . Calculate the maximum friction force. For a two-force body it may be easier to use the resultant force R and the angle f it makes with the normal to the surface. W C G q b A R fs AC GC r 3r/8 sin (180o - q) = sin fs = sin 16.70o C q G sin (180o - q) = 0.76626 180o - q = 50.0o and 130o q = 130o and 50.0o fs fs q = 130o impossible q = 50.0o A