Reference no: EM13810885

1) You are driving along with a furry dice hanging from the ceiling of your car. You observe that the furry dice are motionless relative to the car. Draw a clearly labeled free-body diagram for the furry dice if your car has a uniform velocity. Draw a clearly labeled free-body diagram for the furry dice if your car is speeding up uniformly.

Recall that a free-body diagram (FBD) is a diagram where the body (here furry dice) is represented by a dot and the forces acting on that body are represented by arrows emanating from that dot. What forces are acting on the furry dice?

By "motionless relative to the car" the question is telling you that the furry dice are not swinging around, by "uniform velocity" the question tells you there is no acceleration here and the net force is equal to zero (in both x- and y- directions). Finally, by stating that the "car is speeding up" it now implies that the dice, which recall are "motionless relative to the car", must also be accelerating and for this to happen there must be a net force in the direction of acceleration.

2) A firefighter who weighs 712 N slides down a vertical pole with an acceleration of 3ms^-2, directed downward. What are the magnitude and direction (up or down) of the vertical force on the firefighter from the pole and the magnitude and direction of the vertical force on the pole from the firefighter?

i. Again this is a problem which requires a FBD. Draw a FBD for the firefighter. What forces are acting on him, and what must the net force on the firefighter be for his/her acceleration to be 3ms^-2

ii. Once you've found the force on the firefighter from the pole, the force on the pole from the firefighter is easy, right? Look up Newtons' 3rd law.

3) The coefficient for static friction for rubber on dry asphalt is from 0.35 to 1.2 (average of say 0.775), while for rubber on wet asphalt its from 0.25 to 0.8 (average of say 0.525). These values are taken from Baker, J.S., "Traffic Accident Investigation Manual", 1975.

Consider a car traveling at 20.1168 ms^-1 (45 mph) with a driver reaction time of 0.75 s on a dry road.

At what speed should the driver travel in wet conditions to maintain the same stopping distance? Why is the coefficient of static friction used here? Why not the coefficient of kinetic friction?

i. First find the stopping distance for the car on a dry road. Note that there are two parts to this. First the car moves at a constant velocity for 0.75 s (because the car doesn't decelerate until after the reaction time) and then it decelerates linearly from the initial velocity to the final velocity (zero because it stops). There are different ways to solve this. You could write down an equation for distance with two terms. The first term would be distance during the reaction time is velocity multiplied by time.

The second term would have to take into consideration the deceleration (and the coefficient for static friction). Alternatively, you could sketch a plot of velocity versus time (constant for 0.75 s and then decreasing linearly to zero with a slope equal to deceleration). How do you get distance from a velocity vs time graph?

ii. Repeat what you just did, but in reverse. Now you know the stopping distance (from the first part) and you do the same math (or use the same kind of velocity vs time graph) to find the initial velocity.

iii. Why static and not kinetic friction? Google ABS breaks.