friction (urgent)

AbcDef

2011-10-10 5:58 am

1. A 1000 kg boat is traveling at 90 km/h when its engine is shut off. The magnitude of the frictional force f between boat and water is proportional to the speed v of the boat: f = 70v, where v is in meters per second and f is in newtons. Find the time required for the boat to slow to 45 km/h.
ans: 9.9s

2. An initially stationary box of sand is to be pulled across a floor by means of a cable in which the tension should not exceed 1100 N. The coefficient of static friction between the box and the floor is 0.35. (a) What should be the angle between the cable and the horizontal in order to pull the greatest possible amount of sand, and (b) what is the weight of the sand and box in that situation?
ans: a) 19(degree), b)3.3kN
Can some also explain what's q2 asking? I can't even understand the problem.

thanks

回答 (1)

天同

2011-10-10 7:42 am

✔ 最佳答案

1. Use: force = mass x acceleration
-70v = 1000(dv/dt)
hence, dv/v = -(70/1000)dt
integrate on both sides with v from 90 km/h to 45 km/h, and t from 0 to t
ln(45/90) = -(70/1000)t
i.e. t = (1/0.07).ln(90/45) s = 9.9 s

2. (a) Let m be the mass of the sand and box and a be the angle at which the cable makes with the horizontal.

Hence, horizontally: 1100.cos(a) = 0.35R -------------------- (1)
where R is the normal reaction

Vertically: mg = R + 1100.sin(a)
i.e. R = mg - 1100.sin(a) -------------------------- (2)
where g is the acceleration due to gravity

Substitute (2) into (1),
1100cos(a) = 0.35[mg - 1100sin(a)]
hence, 0.35mg = 1100cos(a) + 385sin(a) ------------------------- (3)
differentiate on bothe sides with respect to angle a,
0.35g(dm/da) = -1100sin(a) + 385cos(a)

since dm/da = 0 at max value of m
0 = -1100sin(a) + 385cos(a)
tan(a) = 385/1100
a = 19.29 degrees

(b) substitute the value of a into (3),
0.35mg = 1100cos(19.29) + 385sin(19.29)
mg = 3330 N = 3.33 kN

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