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Gaz031
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#21
Report 15 years ago
#21
(Original post by shift3)
Gaz, can you do question 35? (review ex p157)
(a)
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.
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shift3
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#22
Report Thread starter 15 years ago
#22
Oh.. The reaction of the rail was directed towards the rod, not to the ground.

Thanks Gaz!
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