What is the frictional force at constant velocity?

AI Thread Summary
When a 10 N force is applied to a cart moving at constant velocity, the frictional force must equal the applied force to maintain equilibrium, resulting in a frictional force of 10 N acting in the opposite direction. Newton's second law indicates that with zero acceleration, the net force is zero, meaning all forces must balance. The discussion clarifies that both static and kinetic friction concepts are relevant, but the key point is that the frictional force matches the applied force when velocity is constant. The forces acting on the cart include weight, normal force, applied force, and friction force, which cancel each other out. Understanding this balance resolves the confusion regarding the frictional force's magnitude.
moonbase
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Homework Statement


A 10 N force is horizontally applied to a cart from the left. The cart moves to the right with a constant velocity. What is the magnitude and direction of the frictional force? (In terms of mass)

Homework Equations


Fnet=ma
fk=ukN

The Attempt at a Solution


I know Newton's second law shouldn't apply since the acceleration equals zero so there is no net force. But I'm confused as to how I can get all the forces to cancel out if the applied force needs to be greater than the frictional force in order for the cart to move. So the frictional force can't be 10 N. Is there a force I'm missing or is there something wrong with my reasoning?
 
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moonbase said:

Homework Statement


A 10 N force is horizontally applied to a cart from the left. The cart moves to the right with a constant velocity. What is the magnitude and direction of the frictional force? (In terms of mass)

Homework Equations


Fnet=ma
fk=ukN

The Attempt at a Solution


I know Newton's second law shouldn't apply since the acceleration equals zero so there is no net force. But I'm confused as to how I can get all the forces to cancel out if the applied force needs to be greater than the frictional force in order for the cart to move. So the frictional force can't be 10 N. Is there are force I'm missing or is there something wrong with my reasoning?

your problem lies in the statement in red, above.

You do, and have, used Newton's second law when you concluded "... so there is no net force".

getting all the forces to cancel out is easy ...

Weight down
Normal Reaction Force up
Applied Force to the right [applied from the left]
Friction Force to the left.

They cancel in pairs.
 
Oh okay, I was getting static and kinetic friction mixed up. Thanks
 
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