Please note that, although correct mumerical answers are impoIl clarity of your

Question

Please note that, although correct mumerical answers are impoIl clarity of your reason Therefore, make certain to steps and any explanatory points II, and such show all of your work, including all intermediate algebra remarks you may deem appropriate. Each problem is worth 30 ing matter as well, and such factors will be taken into Problem 1 A car of mass 1600 kg travels around a level curve of radius 40 m. (a) If the coefficient of static friction between the tires and the road is 0.70, how fast can the car travel without "spinning out?" (b) On a rainy day the coefficient of friction is reduced to 0.30. What is the radius of the tightest curve that can be negotiated safely at the same speed found in part (a)? Problem 2 A motorcycle is traveling up one side of the hill and down the other side. The crest of the hill is a circular arc with a radius of 55.0 m. (a) Starting from Newton's second law (i.e., DO NOT QUOTE RESULTS FROM OTHER PROBLEMS), determine the maximum speed that the cycle can have while moving over the crest without losing contact with the road. (b) If the motorcycle travels across a dip (a circular arc with the same radius as in part (a)) at twice the speed found in part (a), find the ratio between the apparent weight of the motorcycle plus its rider and their true weight. Problem 3 A 46-kg boy sets out to walk along a 25-kg uniform horizontal beam, 5.0 m long. The beam rests upon, but is not anchored to, two supports as shown below. (a) Find the forces exerted by each of the two suppor when the boy initially stands exactly above the support at the left end. The boy now starts walking toward the right end. How close to the right s the boy when the beam is just about to tip over? 3.0 m

Explanation / Answer

1] The centripetal force due to the circular motion should be equal to the frictional force

mv2/r = usmg

=> v2/r = usg

=> v = [0.7 x 9.8 x 40]1/2 = 16.565 m/s.

if the coefficient of static friction was us = 0.3, then

16.565 = [0.3 x 9.8 x r]1/2

=> r = 93.333 m.

this will be the new radius of the track.

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