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Fall 2008 -- Purdue University -- West Lafayette, IN
Blog Post LABELS
- Blog Challenge Questions (9)
- Chapter 2 (2)
- Chapter 3 (19)
- Chapter 5 (11)
- Chapter 6 (23)
- Chapter 7 (7)
- Chapter 8 (9)
- Exam 1 (5)
- Exam 2 (4)
- Exam 3 (17)
- Exams (2)
- Final Exam (7)
- General (6)
- Gravity (2)
- Homework (2)
- in class (1)
- Lectures (1)
- Quizzes (2)
- Spring 2008 Archive (87)
- Spring-Mass Simulator (1)
- Student-Generated Solutions (1)
- Summer 2008 Archive (72)
5 comments:
I understand mathematically how to find the friction force for part b, however I'm having trouble rationalizing it. I know that since the car begins to turn we now have components that are in the et and en direction, where it was only et before the turn.
Why does the friction have both of these components when it opposes the motion of the object? I know that it has to have this other component, but if it was only opposing motion, it would all be in the et direction. Could someone help clarify this for me?
I am pretty sure that the friction force is in the e_n direction because it is keeping the car on the road, but i am still having trouble figuring out the answer. I set the force equal to m*v^2/rho. This still gives me the wrong answer. What am i doing wrong?
Janet: I think that we should really rely on math for explaination to your question. Before the turn, v^2/rho was 0, because rho was nearly infinity (though not exactly because of the undefined case). Therefore, before the turn, both components existed; but e_n was ~0.
If you were going to visualize the friction force for part B, you can draw the actual vector on the e_n & e_t axis. Since it is slowing the car down, its opposing the motion of the car.. but it is also needs to keep the car on the road and points inwards to follow the curved path, as John V said.
Final thought - an object in motion tends to stay in motion unless acted on by another force... because the car is turning, there must be a force allowing the car to do so. Otherwise, the car would continue in a straight line. I hope that helped.. a little..
John: One of the keys to this problem for me was finding v_dot (there's got to be an easier way than how I did it). Also, in my FBDs the friction force was broken into two components in all parts of the problem.
Thanks Jon, that does help some...the FBDs still slightly confuse me though.
Well the I know the friction force is in BOTH the en and et directions. john, you were correct in setting the force equal to v^2/rho; however that's only part of the solution. don't forget et which is v dot . you can find v dot because you know deceleration is constant and you know how far total the car travels.
note that there is no en force for part a and part c. part a is not on a curve so v^2/rho is 0. part c v=0, so v^2/rho is 0. for part b you find en AND et, then take the square root of the squared sums to find the net force.
at point b i believe there are two components to acceleration. one in the n and t. however you need to find velocity the book gives the equation v_b^2=v_a^2 + 2a_t dS . I am wondering if i find this acceleration can i multiply this by mass and then square both a_t and a_n add then together and take the square root.
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