practice with kinetic friction renate fiora. let’s try solving a problem involving kinetic...

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Practice with Kinetic Friction Renate Fiora

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Page 1: Practice with Kinetic Friction Renate Fiora. Let’s try solving a problem involving kinetic friction. Remember the equation for the force of friction:

Practice with Kinetic Friction

Renate Fiora

Page 2: Practice with Kinetic Friction Renate Fiora. Let’s try solving a problem involving kinetic friction. Remember the equation for the force of friction:

• Let’s try solving a problem involving kinetic friction.

• Remember the equation for the force of friction:

Ff = FN

• Don’t forget to include a drawing and free-body diagram.

Page 3: Practice with Kinetic Friction Renate Fiora. Let’s try solving a problem involving kinetic friction. Remember the equation for the force of friction:

Bonnie and Clyde are sliding a 300. kg bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from

behind with 385 N of force while Bonnie pulls forward on a rope with 350 N of force. What is the safe’s coefficient of kinetic friction on the

bank floor?

Try it on your own, then advance to the next slide to see the solution.

Page 4: Practice with Kinetic Friction Renate Fiora. Let’s try solving a problem involving kinetic friction. Remember the equation for the force of friction:

Bonnie and Clyde are sliding a 300. kg bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from

behind with 385 N of force while Bonnie pulls forward on a rope with 350 N of force. What is the safe’s coefficient of kinetic friction on the

bank floor?

Because there’s no net force, we can say:

Fy = FN – FW = 0 FN = FW = mg

Fx = FB + FC – Ff = 0 FB + FC = Ff

So, = Ff /FN = (FB + FC)/mg

= (350N + 385N)/(300kg)(9.8 m/s2)=0.25

a = 0

FW

Ff

FB

FN

FC

+x

+y

FW

FN

Fnet = 0

FBFf

FC

Page 5: Practice with Kinetic Friction Renate Fiora. Let’s try solving a problem involving kinetic friction. Remember the equation for the force of friction:

A man slides a 45 g crate at a constant velocity across a horizontal floor by pulling on a rope attached to the crate. The angle between the

rope and horizontal is 33 and the coefficient of kinetic friction between crate and floor is 0.63. Determine the tension in the rope.

Try it on your own, then advance to the next slide to see the solution.

Page 6: Practice with Kinetic Friction Renate Fiora. Let’s try solving a problem involving kinetic friction. Remember the equation for the force of friction:

A man slides a 45 kg crate at a constant velocity across a horizontal floor by pulling on a rope attached to the crate. The angle between the

rope and horizontal is 33 and the coefficient of kinetic friction between crate and floor is 0.63. Determine the tension in the rope.

There’s no net force, so we can say:Fy = FN + FTy – FW = 0

FN + FTy = FW = mg

FN = mg - FT sin

Fx = FTx – Ff = 0

FT cos = Ff

So, Ff = FN

FT cos = mg - FT sin

FT (cos + sin = mg

FT = mg/(cos + sin

a = 0

FWFf

FT

FN

+x

+y

FW

FN

Fnet = 0

FT

Ff

Now, plug in the values:FT = /(cos + sin

FT =