Calculating Kinetic Friction

Renate Fiora

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

• Remember the equation for the force of friction:

Ff = sFN

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

A 50.0 kg steel file cabinet is in the back of a dump truck. The truck’s bed, also made of steel, is slowly tilted. (a) What is the size of the static

friction force on the cabinet when the bed is tilted at 20? (b) If the coefficient of static friction for steel on steel is 0.80, at what angle will

the file cabinet begin to slide?

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

A 50.0 kg steel file cabinet is in the back of a dump truck. The truck’s bed, also made of steel, is slowly tilted.

y-direction: FN – FWy = 0 FN = FWy

FN = FW cos = mg cos

x-direction: FWx – Ff = 0 FWx = Ff

Ff = FW sin = mg sin

Ff = 50.0kg(9.8m/s2)sin 20

Ff = 168 N

FW

FN

Ff

+x

+y

FW

FN

Fnet = 0

Ff

(a) What is the size of the static friction force on the cabinet when the bed is tilted at 20?

A 50.0 kg steel file cabinet is in the back of a dump truck. The truck’s bed, also made of steel, is slowly tilted.

From part (a),FN = mg cos & Ff = mg sin

The cabinet will slide when the FWx

equals the max force of friction.Ff(max) = sFN = s mg cos

mg sin = s mg cos

s = sin /cos = tan

= tan-1 s = tan-1

= 39

FW

FN

Ff

+x

+y

FW

FN

Fnet = 0

Ff

(b) If the coefficient of static friction for steel on steel is 0.80, at what angle will the file cabinet begin to slide?