Solve Physics Problem: Coefficient of Friction on Steep Incline with Supertruck

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In summary, the coefficient of friction for the supertruck on a steep 65 degree incline with an acceleration of 3.3 m/s^2, a mass of 1000 kg, and an engine applied force of 1.55 X 10^4 Newtons is approximately 0.82. This is calculated by finding the net force and acceleration of the truck, comparing it to the provided acceleration, and using the normal force to determine the coefficient of friction.
  • #1
Charlie Brown
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A supertruck is fighting its way up an steep 65 degree incline with an acceleration of 3.3 m/s^2. If the truck has a mass=1.00 tonne and an engine applied force of 1.55 X 10^4 Newtons, find the co-efficient of friction.

1 tonne = 1000 kg

**I got 0.80 as my answer...can anyone confirm this?
 
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  • #2
Find the magnitude of gravity down the slope
Find the ideal acceleration of the truck by dividing force and mass.
Find a net force from the above two forces.
Find a net acceleration from this net force. (Divide by mass)
Compare this net acceleration to the acceleration provided.
Find the friction force by subtracting the two forces.
Divide by the normal force to get coefficient of friction.
Normal force is [tex]mgcos(\vartheta)[/tex]
 
  • #3
whozum said:
Find the magnitude of gravity down the slope
Find the ideal acceleration of the truck by dividing force and mass.
Find a net force from the above two forces.
Find a net acceleration from this net force. (Divide by mass)
Compare this net acceleration to the acceleration provided.
Find the friction force by subtracting the two forces.
Divide by the normal force to get coefficient of friction.
Normal force is [tex]mgcos(\vartheta)[/tex]

Thank you...is 0.801 the answer?
 
  • #4
Resultant force = ma

let friction = F.

so, resolving forces parallel to the slope, using g=9.81N/kg:
F = ma
1.55*10^4 - F - 1000gcos65 = 1000*3.3
F= 8054 N

F=uR, where u = coefficient of friction
u = F/R
u = 8054/(1000*9.81)
u = 0.82
 
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