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Published byLorraine Hudson Modified over 8 years ago
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Calculating Kinetic Friction Renate Fiora
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Let’s try solving a problem involving static friction. Remember the equation for the force of friction: F f = s F N Don’t forget to include a drawing and free- body diagram.
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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.
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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: F N – F Wy = 0 F N = F Wy F N = F W cos = mg cos x-direction: F Wx – F f = 0 F Wx = F f F f = F W sin = mg sin F f = 50.0kg(9.8m/s 2 )sin 20 F f = 168 N FWFW FNFN FfFf +x +y FWFW FNFN F net = 0 FfFf (a) What is the size of the static friction force on the cabinet when the bed is tilted at 20 ?
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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), F N = mg cos & F f = mg sin The cabinet will slide when the F Wx equals the max force of friction. F f(max) = s F N = s mg cos mg sin = s mg cos s = sin /cos = tan = tan -1 s = tan -1 = 39 FWFW FNFN FfFf +x +y FWFW FNFN F net = 0 FfFf (b) If the coefficient of static friction for steel on steel is 0.80, at what angle will the file cabinet begin to slide?
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