How big is the friction force on the truck?

AI Thread Summary
To calculate the friction force on a truck parked on a 20-degree slope, the correct approach involves using the weight of the truck and the sine of the incline angle. The truck's weight is 4200 kg, resulting in a gravitational force of approximately 412.2 N acting down the slope when multiplied by sin(20). The coefficient of static friction of 0.90 indicates the maximum friction force, which is not directly used in the final calculation but serves to understand the limits of static friction. The static friction force must balance the component of gravitational force acting down the slope, confirming that the truck is in equilibrium. The final answer for the friction force is determined to be approximately 412.2 N.
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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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