Kinetic Friction of sliding down a pole

[lacar213]

Homework Statement

A firefighter whose weight is 812 N is sliding down a vertical pole, her speed increasing at the rate of 1.45 m/s2. Gravity and friction are the two significant forces acting on her. What is the magnitude of the frictional force?

Homework Equations

Fk = ukFn
uk = coefficient of kinetic friction

The Attempt at a Solution

I'm not sure what the coefficient of KF should be so I just tried ..
812*1.45 = 1177.4
I'm not sure how to incorporate the gravity and friction into the equation.

 

[Doc Al]

I'm not sure what the coefficient of KF should be so I just tried ..

You're not given the coefficient of friction. Luckily you won't need it--you'll figure out the friction force another way.

812*1.45 = 1177.4
I'm not sure how to incorporate the gravity and friction into the equation.

Gravity and friction are the two forces acting on the firefighter. You are given her acceleration. Apply Newton's 2nd law: ∑ F = ma

(What's the firefighter's mass?)

 

[lacar213]

OK ...
F = ma
mass -> 812 N/9.8 = 82.8571 kg
F = (82.8571)(1.45)
F = 120.143 N

Is it that simple?

 

[Doc Al]

F = ma
mass -> 812 N/9.8 = 82.8571 kg

Good.

F = (82.8571)(1.45)
F = 120.143 N

Is it that simple?

Not that simple, but almost. The "F" in F = ma stands for the net force acting on the object. Express the net force in terms of the actual forces acting on the firefighter (weight and friction) by adding those forces up. (∑ F means the sum of the forces.) Then you can solve for the unknown friction force.

Hint: Careful with signs.

 

[lacar213]

Would the next step be to multiple her weight and gravity? Then add that net force to 120.143 N?

F = (82.8571)(9.8) = 812 N - but that's the weight already given??
120+812 = 932.143