How Can a Cart Maintain Constant Velocity Down a Slope?

[phy124]

Homework Statement

Would constant velocity down a slope be achieved by a cart, by having the frictional force equal to that of Wsinθ (Where W is the weight force and θ is the angle between the slope and the horizontal).

Homework Equations

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The Attempt at a Solution

For example if θ was equal to 15 and W was equal to 19.

Then to achieve a constant velocity the frictional force would need to equal approximately 4.92 N.

Or am I going in the wrong direction with this?

Any help is appreciated and please excuse any ambiguity within this post.

 

[tiny-tim]

welcome to pf!

hi phy124! welcome to pf!

Would constant velocity down a slope be achieved by a cart, by having the frictional force equal to that of Wsinθ (Where W is the weight force and θ is the angle between the slope and the horizontal).

if the brakes are on, yes …

good ol' Newton's second law in that direction means that ∑F = 0, so friction = Wsinθ …

what is worrying you about that? ​

 

[phy124]

hi phy124! welcome to pf!

Thank you tiny tim.

if the brakes are on, yes …

good ol' Newton's second law in that direction means that ∑F = 0, so friction = Wsinθ …

what is worrying you about that? ​

Yes well here lies the problem, the cart doesn't have brakes and I'm pretty sure that friction acts in the opposite direction to that of the way which the wheels are moving, so it is in the same direction as the carts movement and thus making Fr = Wsinθ wrong?

 

[tiny-tim]

hi phy124!

(just got up :zzz: …)

… the cart doesn't have brakes and I'm pretty sure that friction acts in the opposite direction to that of the way which the wheels are moving, so it is in the same direction as the carts movement and thus making Fr = Wsinθ wrong?

ah, good point …

yes, if there's wheels and no brakes, then the friction with the ground is irrelevant (there is no work done ) …

the only thing slowing the cart is friction in the bearings (between the wheel and the axle), and air resistance …

usually not enough to overcome gravity and keep a constant velocity!

(actually you're wrong about the direction of friction from the ground … for driving or braking wheels, attached to the engine, the friction is in the same direction as the acceleration or braking, but for non-driving non-braking wheels, surprisingly it's in the opposite direction! )

 

[9une]

Hi,
The cart is given three forces: gravity(W), support force(N) and frictional resisntance(F). If the cart is moving with constant velocity along the slope, the force along the slope should be zero, which means Wsinθ = F. As for dynamic frictional resistance, we have the equation F= Nμ, where μ is friction coefficient. Another equation we have is N = Wcosθ. So the final equation to satify your requirement is tanθ = μ. Without information about the friction coefficient, we could get the answer.

 

[tiny-tim]

welcome to pf!

hi 9une! welcome to pf!

… So the final equation to satify your requirement is tanθ = μ.

no, as phy124 points out, µ is irrelevant if the cart has wheels and no brakes

 

[9une]

no, as phy124 points out, µ is irrelevant if the cart has wheels and no brakes [/QUOTE]

In my opinion, µ is decided by wheels and brakes. How could we say µ is irrelevant?

 

[tiny-tim]

In my opinion, µ is decided by wheels and brakes. How could we say µ is irrelevant?

wake up! :zzz: have some coffee! …

… the cart doesn't have brakes …

 

[phy124]

hi phy124!

(just got up :zzz: …)


ah, good point …

yes, if there's wheels and no brakes, then the friction with the ground is irrelevant (there is no work done ) …

the only thing slowing the cart is friction in the bearings (between the wheel and the axle), and air resistance …

usually not enough to overcome gravity and keep a constant velocity!

(actually you're wrong about the direction of friction from the ground … for driving or braking wheels, attached to the engine, the friction is in the same direction as the acceleration or braking, but for non-driving non-braking wheels, surprisingly it's in the opposite direction! )

Ah I see, thank you for your insight, you were very helpful. I'll be sure to come back here if I have any problems in the future ;)