Friction and Velocity: Solving for Constant Velocity on an Inclined Surface

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To achieve constant velocity while sliding down an inclined surface, the forces acting on the car must balance, resulting in zero acceleration. The gravitational force component parallel to the slope is expressed as mg sin(θ), while the frictional force is μmg cos(θ). Setting the net force equation to zero (ma = 0) leads to the relationship mg sin(θ) = μmg cos(θ). This simplifies to tan(θ) = μ, allowing for the calculation of the inclination angle θ. The correct approach involves understanding the distinction between the parallel and perpendicular components of weight in relation to friction.
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A car of mass 1000kg is sliding down a hill. The coefficients of friction between the cars tires and the ground are u=0.89 and u[k]=.61. For what inclination angle will the car slide down the hill with a constant velocity.
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So what I figured was that the car will have constant velocity when there is 0 acceleration. So F-f[friction]=0 ( which I don't know is correct or not). Then I set ma=u[k]mgcosQ, and solved for cosQ to get arccos(a/u[k]g). I just don't think any of what I'm doing is correct. Where have I gone wrong?
 
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mohabitar said:
So what I figured was that the car will have constant velocity when there is 0 acceleration. So F-f[friction]=0 ( which I don't know is correct or not). Then I set ma=u[k]mgcosQ, and solved for cosQ to get arccos(a/u[k]g). I just don't think any of what I'm doing is correct. Where have I gone wrong?

well firstly, you need to get the component of the weight parallel to the slope, that will be your 'F'. When you get that you can solve it the way you were solving before.
 
Well I thought that's what I did, no? Thats what mgcosQ is, the parallel component of the weight..
 
mohabitar said:
Well I thought that's what I did, no? Thats what mgcosQ is, the parallel component of the weight..

Nope, remember that friction =μN = μmgcosθ, which you correctly identified and this is normal to the slope, so there is only one other component left, which is parallel.
 
Ok so I have two things here: mgcosQ and mgsinQ. So mgcosQ is perpendicular weight, and mgsinQ is the parallel weight? So then what? ma=uk(mgsinQ-mgcosQ)?
 
mohabitar said:
Ok so I have two things here: mgcosQ and mgsinQ. So mgcosQ is perpendicular weight, and mgsinQ is the parallel weight? So then what? ma=uk(mgsinQ-mgcosQ)?

No you will have ma=mgsinθ - μmgcosθ and you said that for constant velocity, a = 0, so θ is?
 

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