A city planner is working on the redesign of a hilly portion of acity. An import

Question

A city planner is working on the redesign of a hilly portion of acity. An important consideration is how steep the roads can be sothat even low-powered cars can get up the hills without slowingdown. A particular small car, with a mass of 920 s . Using these data, calculate the maximum steepness of a hill. Find theta. I haven't a clue how to solve this. We discussed the problem inclass and the formula my professor gave uswas F-mgcos theta but I have no idea howto use this. Will rate lifesaver since this is due at midnight. km/h ) in 13.7 m/s (79 kg , can accelerate on a levelroad from rest to 22

Explanation / Answer

the idea is that the max force the car can apply on levelground is found by first finding its acceleration .      a = change in speed /time = 22 / 13.7 =   1.606m/s2 . so the force it can apply is    F = ma = 920 * 1.606 =    1477.4 Newtons . This depends on the contact the car makes with the road...i.e. the friction force. We can write: .           Force applied = coeff of friction *normal   or          F = u n . On level ground, the normal force is equal to the weight ofthe car. But on an incline, the normal force is the weight of thecar times cos   (this is what your prof was tryingto direct you toward...) . So...    now we can find the force the carcan apply to pull itself up the incline: .           levelground:           Flevel =   u mg .          incline:          Fincline = u mg cos . Divide the equationsand          Fincline / Flevel   = cos     or .         Fincline   =  Flevel cos . Now... not only does the force the car pulls with decreasewhile on the incline... it ALSO has to fight against gravity! . Force of gravity pulling down incline =   mgsin       (note: when angle iszero, this force is zero... as it should be) . So finally... for the car to be able to drive up the hill, theforce the car can pull with must at least equal the force gravitypulls back with. Or... .            Fincline =   mgsin                Flevel cos  = mgsin . Divide both sides bycos                Flevel   = mg tan .    tan = Flevel / mg =   1477.4 / 920 * 9.80  =   0.16386 .             = arctan (0.16386) =    9.31 degrees is the maxangle         wow! yeah,this was a sneaky problem . so the force it can apply is    F = ma = 920 * 1.606 =    1477.4 Newtons . This depends on the contact the car makes with the road...i.e. the friction force. We can write: .           Force applied = coeff of friction *normal   or          F = u n . On level ground, the normal force is equal to the weight ofthe car. But on an incline, the normal force is the weight of thecar times cos   (this is what your prof was tryingto direct you toward...) . So...    now we can find the force the carcan apply to pull itself up the incline: .           levelground:           Flevel =   u mg .          incline:          Fincline = u mg cos . Divide the equationsand          Fincline / Flevel   = cos     or .         Fincline   =  Flevel cos . Now... not only does the force the car pulls with decreasewhile on the incline... it ALSO has to fight against gravity! . Force of gravity pulling down incline =   mgsin       (note: when angle iszero, this force is zero... as it should be) . So finally... for the car to be able to drive up the hill, theforce the car can pull with must at least equal the force gravitypulls back with. Or... .            Fincline =   mgsin                Flevel cos  = mgsin . Divide both sides bycos                Flevel   = mg tan .    tan = Flevel / mg =   1477.4 / 920 * 9.80  =   0.16386 .             = arctan (0.16386) =    9.31 degrees is the maxangle         wow! yeah,this was a sneaky problem

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.