Gaz, can you do question 35? (review ex p157)
As the rail is smooth the reaction acts only perpendicular to the point of contact.
Let the reaction at the rail be S and the reaction at the ground be R and F.
Let the angle of elevation be t.
As the reaction is pependicular to the slope it acts at an angle (90-t) to the horizontal.
Resolving vertically: Ssin(90-t) + R = 20g
Scost + R = 20g
Resolving horizontally: Scos(90-t) = F
Ssint = F
Taking moments about A:
C = AC
(20g)(3cost) = (Scost)(5cost) + (Ssint)(5sint)
60gcost = 5S(sin^2 t + cos^2 t)
S = 60gcost/5
Using trigonometry. sint = 3/5. Hence cost = 4/5
Substituting gives S = 94N(nearest N)
(b) F = uR
u = F/R = (Ssint)/(20g-Scost) [From earlier resolving force equations]
Substituting exact values gives 0.47 (2dp)
(c) R = 20g-Scost = 120.736
F = Ssint = 56.448
Magnitude = rt[120.736^2 + 56.448^2] = 133N (nearest N)
Feel free to ask if there's anything else i can help with.