As a city planner, you receive complaints from local residents about the safety

Question

As a city planner, you receive complaints from local residents about the safety of nearby roads and streets. One complaint concerns a stop sign at the corner of Pine Street and 1st Street. Residents complain that the speed limit in the area (89 km/h) is too high to allow vehicles to stop in time. Under normal conditions this is not a problem, but when fog rolls in visibility can reduce to only 47 meters. Since fog is a common occurrence in this region, you decide to investigate. The state highway department states that the effective coefficient of friction between a rolling wheel and asphalt ranges between 0.689 and 0.770, whereas the effective coefficient of friction between a skidding (locked) wheel and asphalt ranges between 0.450 and 0.617. Vehicles of all types travel on the road, from small VW bugs with a mass of 651 kg to large trucks with mass 3775 kg. Considering that some drivers will brake properly when slowing down and others will skid to stop, calculate the minimum and maximum braking distance needed to ensure that all vehicles traveling at the posted speed limit can stop before reaching the intersection. Given that the goal is to allow all vehicles to come safely to a stop before reaching the intersection, calculate the maximum desired speed limit. A heavy sled is being pulled by two people as shown in the figure. The coefficient of static friction between the sled and the ground is mu_s = 0.635, and the kinetic friction coefficient is mu_k = 0.459. The combined mass of the sled and its load is m = 276 kg. The ropes are separated by an angle Phi = 28 degree, and they make an angle theta = 32.4 degree with the horizontal. Assuming both ropes pull equally hard, what is the minimum rope tension required to get the sled moving? If this rope tension is maintained after the sled starts moving, what is the sled's acceleration?

Explanation / Answer

Fn = mg - 2Fsintheta = 276kg * 9.8m/s² - 2 * T * sin32.4

Fn = 2704.8 - 1.072*T

static friction Ff = mu*Fn = 0.635 * (2704.8 - 1.072T) = 1717.548N - 0.681T

The horizontal component in the direction of motion of the two ropes must overcome the friction:

2 * T * costheta * cos(phi/2) = 1717.548 N - 0.681T

1.638 T = 1717.548 N - 0.681*T

T = 740.64 N

The friction WAS Ff = 1717.548 - 0.681*740.64N = 1213.172 N

so that IS the applied horizontal force.

The friction NOW IS Ff = 0.459 * (2704.8 - 1.072*740.64N) = 877.07 N

so the net force

Fnet = 1213.172 - 877.07 = 336.1 N

acceleration a = Fnet / mass = 336.1/ 276kg = 1.22 m/s^2

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