## brinethery 3 years ago The coefficient of static friction between the crate and the bed of the truck is mew_s = .6 (I just made this up because his answer key is done in general terms). The coefficient of kinetic friction between the crate and the bed of the truck is mew_k = .3. The truck will go around a corner with a radius of curvature of r=41 meters. For safety, the truck driver must be able to turn and slow down at the same time. The truck must slow down at a rate of 1.6 m/s^2 while staying on the road. Now what is the maximum speed that the truck can go around the corner so that it can slow down at this rate without causing the box to slide on the bed of the truck?

1. furnessj

Does the crate have a mass?

2. brinethery

The mass is irrelevant because it ends up canceling when you sum the forces in the x-direction, which is why he didn't give it. My teacher is very hard in that the homework assignments are deceivingly easy and then the quiz questions throw all of us off. Oh and the 1.6 m/s^2 is actually 1.6 m/s. Sorry about that. Someone in the chatbox asked me to post it and I was really tired as I was typing.

3. furnessj

It is including the 1.6 that I don't get really, but if I equate the limiting frictional force with the centripetal force I get tangential velocity as 15.53 m/s. Not sure what the 1.6 m/s actually is, a rate of 1.6m/s sounds like a deceleration (m/s/s were right?). I think my head has gone on this for the day!

4. brinethery

I can scan in the answer key for you. Again, someone asked me to post it after I told them how frustrated I was with this question. The answer key was done in general terms, but you'll get the idea.

5. furnessj

you can certainly do that if it is still annoying you, the full question should suffice but answers to work towards are always useful!

6. brinethery

Actually I figured it out. You have tangential acceleration which is tangent to the circle. That's going in the opposite direction of the car. You first have to find the maximum velocity the car can go without the box sliding (not taking into account the tangential acceleration). Summation of forces_x = mv^2/r = m*g*mew_s Summation of forces_y = 0 = -mg + mg We don't need summation in y, I just wrote it to be complete. So mass cancels like Promised. v^2/r = g*mew_s v = sqrt (r*g*mew_s) Now, find the acceleration. a = v^2/r This acceleration will be the resultant vector (hypotenuse) You know what the tangential acceleration is and what the resultant vector is, so you just need to use the pythagorean theorem to find what "b" is b = sqrt (c2 - a^2), where c is the resultant vector and a is the tangential vector. The "b" value is going to be the new centripetal acceleration.

7. brinethery

**c^2 on that 2nd to last line.

8. Moyo30

Hi juss been lookin at your quest. It requires the maximum velocity to be found that will allow the truck to slow down at 1.6m/s^2 without moving the crate. First part you equated centripital force to force req to move the box. thus found an expression for max centripetal velocity. I don't get the second part.

9. Moyo30

can't we find the tangetal velocity then find the magnitude of the velocity

10. Moyo30

the other question is the -1.6m/s^2 is it tangental acceleration?

11. brinethery

I'll post the answer key when I'm done with my physics midterm tomorrow. Anyway, the tangential acceleration vector + centripetal acceleration vector = acceleration vector (that was calculated with max velocity the car can go around without sliding).

12. brinethery

Actually, I change my mind. I'm sick of studying! Here one sec I'm scanning it.

13. Moyo30

thx, I have end of semesters tomorrow. it'd help me a long way

14. brinethery

15. brinethery

In the first part, mew_s and the radius are given. With that, you must solve for maximum velocity. For part 2, mew_s, the radius, and a negative acceleration (slowing down) are given. You must first find the new centripetal acceleration. With that, you have your formula a = v^2/r and then you can manipulate that around and solve for your new velocity.

16. Moyo30

Thx so for the first part there is only the centripetal force acting we don't have to worry about the tangential force

17. brinethery

Tangential acceleration on my quiz was 1.6 m/s^2, but slowing down so negative. That was given on the second part, but it's not clear on his answer key. Nothing is clear in the second part, it took me over an hour to figure out what he was trying to say lol!

18. brinethery

Yeah first part is only centripetal.

19. Moyo30

thanks, for the first part the truck is simply going around a bend at constant speed. For the sec part it is slowing down whilst going round the bend. So in addition to the centripetal force and velocity u have decceleration. The resultant acceleration is the vector result of the centripetal and tangental acceleration. SO they used that to get the new velocity

20. brinethery

Yup.

21. Moyo30

Thanks for the question

22. brinethery

Sorry I had a typo in the beginning of my post. It said 1.6 m/s and I fixed and made it 1.6 m/s^2. I always hate it when posters confuse ME, and now it's me doing it to everyone else. Sorry about that :-)

23. brinethery

Hey I have another really good one for you to look at. It's a youtube vid, one sec let me grab it.

24. brinethery

He did this so perfectly: http://www.youtube.com/watch?v=L9StuVjWwaU

25. brinethery

This is WITH friction, so it's a bit trickier. But he really simplifies things.

26. Moyo30

Thx, having a look at it now

27. brinethery

I have this book and it seems to be helping. http://www.amazon.com/Schaums-Solved-Problems-Physics-Outline/dp/0071763465

28. Moyo30

Thx will have a look at it