How big is the friction force on the truck?

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Homework Help Overview

The discussion revolves around calculating the friction force acting on a truck parked on a 20-degree incline. The truck has a mass of 4200 kg, and the coefficient of static friction between the tires and the road is given as 0.90. Participants are exploring the implications of these values in the context of static equilibrium.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of the friction force and its relationship to the incline angle. There are attempts to apply the coefficient of static friction in various ways, leading to confusion about the correct approach. Questions arise regarding the role of static friction and the forces acting on the truck, including the need for a free body diagram.

Discussion Status

The discussion is active, with participants providing guidance on the importance of understanding the forces at play, particularly in relation to static equilibrium. Some participants suggest drawing a free body diagram to clarify the forces involved, while others emphasize the need to consider the direction of friction relative to gravity.

Contextual Notes

There is a mention of the maximum static friction force and its implications for the truck's equilibrium. Participants are also questioning the relevance of the static coefficient in the problem, indicating potential confusion about its application.

chicagobears34
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Calculating friction force on a truck on an incline?

Homework Statement


A 4200kg truck is parked on a 20 degree slope.
How big is the friction force on the truck? The coefficient of static friction between the tires and the road is 0.90.

Homework Equations


F=ma
Ff=mu*m*a

The Attempt at a Solution


Friction Force on the truck is (4200kg*9.81m/s^2 * .90) = 37081.8N
since it's on an incline I am supposed to do 37081.8 * sin20?
so the answer would be 12682.7N but that is wrong
I also tried doing 37081.8 * cos20 which is 34845.5N which is also wrong
Please help me! Idk what I am doing wrong
 
Last edited:
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Please post the whole problem statement.
 
sorry about that. I posted the whole problem now.
 
Draw a free body diagram and the forces acting on the truck.
 
What you found is the max static friction meaning that you would need a force greater than that friction to move the truck. If you were to draw out a FBD and put the friction you had found in you'll see that it's greater than the gravity force acting on the truck down the incline. But you know that this can't be true since the truck is in equilibrium.
 
I did draw a FBD, but I really suck at physics. Wouldn't there just be a force going down on the truck which would be (4200*9.81)cos20? where would the friction point?
can you point me in the right direction?
 
No it would be sin20. Frictions always point against motion so it will be pointing opposite of the gravity force down the incline. Don't forget that a=0 either.
 
alright thanks a bunch Maiq. What was the point of the static coefficient in that problem?
The answer was just (4200*9.81)sin20
 
The static friction does not mean a definite magnitude. It is the force necessary to keep a body in rest, but it can not exceed (normal force) times (coefficient of static friction).

ehild
 
  • #10
They probably put it in there to throw you off or maybe just to show what ehild said.
 

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