1. 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

2. 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.

3. Problem 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 = 16.70o
b = 16.70o 3 8
Geometry:
GC = r (see fig. 5.21)
AC = r

4. sin (180o - q) sin fs AC GC = sin (180o - q) = sin fs = sin 16.70o
Problem Solution
fs = tan = 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

5. sin (180o - q) = sin fs = sin 16.70o sin (180o - q) = 0.76626
Problem 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) =
180o - q = 50.0o and 130o
q = 130o and 50.0o fs fs
q = 130o impossible q = 50.0o A