![]() ![]() ![]() So we can start using the coefficient of kinetic friction It's the ratio between that and the magnitude of the force of contact between this block and the floor or ground here And the magnitude of that force of contact is the same thing as the normal force that the ground is applying on the block the magnitude of the normal force the ground is applying on the block Then once its moving then we can say that this is going to be-this will then be equal to this over here will be equal to the force of friction So this is the force that really overcome friction and this over here will be equal to the force of friction The magnitude of the force of friction over the force of contact the contact force between those two, so over the normal force and it makes sense that the larger the contact force the more that these are being pressed together the little at the atomic level, they kind of really get into each others grooves the more budging force you would need or the more friction force would go against your motion And in either situation the force of friction is going against your motion So even if you push it in that way sounds like force of friction is all of a sudden going to help you So let's think about what the necessary force will we need to overcome the force of friction right here in the static situation So the force of gravity on this block is going to be the gravitational field which is 9.8 m/s^2 times 5 kilograms 9.8 m/s times 5 kilograms gives 49 kilogram meters per second or 49 newtons down This is the force, the magnitude of the force due to gravity the direction is straight down towards the center of the earth The normal force, and that force is there because this block is not accelerating downwards So there must be some force that completely balances off the force of gravity And in this example, it is the normal force So it is acting 49 newtons upward and so these net out. So I have got this block of wood here that has a mass of 5 kilograms and it is sitting on some dirt and we are near the surface of the earth and the coefficient of static friction between this type of wood and this type of dirt is 0.60 and the coefficient of kinetic friction between this type of wood and this type of dirt is 0.55 This was measured by someone else long ago or you found it in some type of a book someplace And let's say we push on this side of the block with a force of a 100 N What is going to happen? So the first thing you might realize is if there is no friction if this was a completely frictionless boundary and there is no air resistance, we are assuming that there is no air resistance in this example That in this dimension, in the horizontal dimension there would only be one force here, this 100 N force It would be completely unbalanced and that would be the net force and so you would have a force going in that direction of a 100 N on a mass of 5 kilograms Force = Mass times acceleration acceleration and force are vector quantities So you would have the force divided by the mass would give you 20 meters per second of acceleration in the rightward direction That is if there were no friction but there is friction in this situation So let's think about how we'll deal with it So the coefficient of friction tells us So this right here is the ratio between the magnitude of the force that I have called the budging force The amount of force you need to apply to get this thing to budge to get this thing to start moving. ![]() A ∥ = Σ F ∥ m (use Newton’s second law for the parallel direction) a ∥ = m g sin θ − F k m (plug in the parallel forces) a ∥ = m g sin θ − μ k F N m (plug in the formula for the force of kinetic friction) a ∥ = m g sin θ − μ k ( m g cos θ ) m (plug in m g cos θ for the normal force F N ) a ∥ = m g sin θ − μ k ( m g cos θ ) m (cancel the mass that’s in the numerator and denominator) a ∥ = g sin θ − μ k ( g cos θ ) (savor the awe when you realize the acceleration doesn’t depend on mass) a ∥ = ( 9.8 m s 2 ) sin 3 0 ∘ − ( 0.150 ) ( 9.8 m s 2 ) cos 3 0 ∘ (plug in numerical values) a ∥ = 3.63 m s 2 (calculate and celebrate) \begin 0 0 0 0 sin θ cos θ sin θ tan θ θ θ θ = m g sin θ − μ s F N (plug in formula for maximum static friction force ) = m g sin θ − μ s ( m g cos θ ) (plug in expression for normal force on an incline ) = m g sin θ − μ s ( m g cos θ ) (divide both sides by m g ) = sin θ − μ s ( cos θ ) (savor the awe when you realize the angle doesn’t depend on the car’s mass) = μ s ( cos θ ) (solve for sin θ ) = μ s (divide both sides by cos θ ) = μ s (replace cos θ sin θ with tan θ ) = tan − 1 ( μ s ) (take inverse tangent of both sides) = tan − 1 ( 0. ![]()
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