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Free Body Diagrams

Use the Free-body diagrams to answer the following:

1. You are driving along the highway, when the car in front of you makes a right-hand turn. What is the best free-body diagram of forces on the turning car, from your point of view behind the car?

2. The speed of the car in front of you is a constant 25m/s. The mass of the car is 800kg. The radius of the turn it is making is 75 meters. What is the magnitude of the resultant force on the car while making the turn?

3. The road is level. What is the magnitude of the normal force of the road on the car as it is making the turn?

4. The coefficient of static friction between the roads and the tires is 0.9. For the same car making the same turn as in the question above, what was the force of static friction on the car while making the turn?

5. For the same car as above, what is the fastest speed possible it can make a turn of the same radius?

6. Suppose the same car, moving at a constant speed of 30 m/s, makes the sharpest turn that it can. What answer is closest to the radius of the turn?

7. The reason the car tried to make a sharp turn is to avoid going over a cliff. Oooop! The car goes off the cliff. What is it's free-body diagram after going off the cliff? (Ignore air resistance).

8. You are looking at a satellite in orbit about the Earth (above the atmosphere). The Earth is to the left of the satellite. What is the free-body diagram of the satellite?

9. The satellite is in a circular orbit, at a distance of 7000 kilometers from the center of the Earth. The mass of the satellite is 450 kg, and its orbital speed is 7560 meters/second. What is the magnitude of the force of the Earth's gravity pulling on the satellite?

10. What is the acceleration (due to gravity) of the satellite in orbit? (in meters/second^2).


Solution Preview

Question #1: E Assuming constant speed, friction force is directed toward the center of
the circular path of the car.
Question #2: C The equation expressing newton's second law is: friction force F= M a.
centripetal acceleration is a= v^2/R, so F= M v^2/R = 6667 nt.
Question #3: B The normal force equals the weight; N= Mg since vertical acceleration is

Solution Summary

Solution contains brief explanations and calculations for each question.