How Can a Cart Maintain Constant Velocity Down a Slope?

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Homework Help Overview

The discussion revolves around the conditions under which a cart can maintain a constant velocity while moving down a slope. Participants explore the relationship between gravitational forces, friction, and the mechanics of motion on an incline.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the balance of forces acting on the cart, particularly the frictional force and gravitational component along the slope. Questions arise regarding the role of friction when the cart lacks brakes and how this affects the motion.

Discussion Status

There is an ongoing exploration of different interpretations of friction's role in the scenario. Some participants suggest that friction is irrelevant without brakes, while others challenge this view, indicating a divergence in understanding. Helpful insights have been shared regarding the mechanics involved, but no consensus has been reached.

Contextual Notes

Participants note the absence of specific information regarding the friction coefficient and the implications of having wheels without brakes on the cart's motion. This lack of information is acknowledged as a constraint in fully resolving the discussion.

phy124
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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.
 
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welcome to pf!

hi phy124! welcome to pf! :smile:
phy124 said:
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? :wink:​
 


tiny-tim said:
hi phy124! welcome to pf! :smile:
Thank you tiny tim.
tiny-tim said:
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? :wink:​

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?
 
hi phy124! :smile:

(just got up :zzz: …)
phy124 said:
… 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 :smile: …

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! :wink:

(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! :rolleyes:)
 
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.
 
welcome to pf!

hi 9une! welcome to pf! :smile:
9une said:
… 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 :wink:
 


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

In my opinion, µ is decided by wheels and brakes. How could we say µ is irrelevant?
 
9une said:
In my opinion, µ is decided by wheels and brakes. How could we say µ is irrelevant?

wake up! :zzz: have some coffee! :smile: …
phy124 said:
… the cart doesn't have brakes …

:wink:
 
tiny-tim said:
hi phy124! :smile:

(just got up :zzz: …)


ah, good point :smile: …

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! :wink:

(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! :rolleyes:)

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 ;)
 

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