A truck tows a trailer of mass 1200 kg at a speed of 12 m/s on a level road.The tension in the coupling is 1000N.Find the tension in the coupling when the truck ascends a raod having inclination of sin^-1(1/6).Assume that the frictional resistance on the inclined plane is the same as that on level road.Please explain it to me.
Asked by Samridh Shukla | 8th Oct, 2010, 10:17: AM
The truck moves at a constant speed of 12 m/s. therefore acceleration = 0.
Making a force balance:
T - μmg = 0
1000 - μ (1200 x 10) = 0
μ = 1000/12000 = 1/12 = 0.083
Now, for the inclination of Θ = sin-1 (1/6) or sinΘ = (1/6), at a constant speed (acceleration = 0):
Force balance gives:
T - (friction force + component of mg parallel to incline) = 0
T - (μmg cosΘ) - mg sinΘ = 0
T = (μmg cosΘ) + mg sinΘ
= (0.083*1200*10*√(35) / 6 ) + (1200*10*1/6) [cosΘ = √(35) / 6 for sinΘ = 1/6]
= 2982 N
Hope that satisfies your query.
Answered by | 8th Oct, 2010, 06:50: PM
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