Force of Friction on object on an incline

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Homework Help Overview

The discussion revolves around calculating the friction force acting on a truck parked on an incline of 18.1 degrees. The original poster presents their calculation of the weight of the truck and attempts to relate it to the friction force using trigonometric functions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between the forces acting on the truck, particularly the normal force and how it relates to friction. There are questions about the correct setup of the triangle used for calculations and the role of the angle in determining the components of gravitational force.

Discussion Status

Some participants provide insights into the nature of friction and its dependence on the normal force, while others express confusion regarding the calculations and the physical principles involved. There is an ongoing exploration of how to correctly apply the concepts of friction and gravity on an incline.

Contextual Notes

Participants note the absence of the coefficient of friction (μ) and discuss whether it is necessary for the problem. The original poster questions the setup of their calculations, indicating a potential misunderstanding of the forces involved.

tascja
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Homework Statement


A 3812.0 kg truck is parked on a 18.1° slope. What is the friction force on the truck?


Homework Equations



The Attempt at a Solution


So i set up a triangle with and then used:
weight of truck= (9.8 m/s^2 *3812 Kg) = 37367.6 N

sin(18.1) = (37367.6 N) /(-Ff)
Ff = -120246 N

Could someone explain why this is wrong.. I am thinking maybe i set up my triangle incorrectly, but i don't know how else i would draw it.. thanks in advance
 
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Friction is based on the force normal to the incline, not normal to the horizontal.
 
I thought friction went parallel to the incline?
 
sorry but I am not following what your saying.. i don't know the materials so i can't find μ...
 
You don't need μ in this case. The Frictional Force is what it balancing the truck's component of gravity down the incline. They just want you to find it. Not calculate it from the Normal force.
 
ok so Ff = mgcos(theta) .. and then is 18.1 my angle?
 
tascja said:
ok so Ff = mgcos(theta) .. and then is 18.1 my angle?

Is M*g*cosθ the component of gravity || to the incline?
 
oh no then i would be mgsinθ = (3812 kg)(9.81 m/s^3)(sin18.1) = 1.162e4 N
 

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