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.

Dear Student

 

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.

 

Team 

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