UCM from AP B Exam (1977 B2) : Confused with solution to #3

In summary, the conversation discusses a box of mass M on a truck traveling at a constant speed v on a circular roadway with radius of curvature R. On the diagram, the forces acting on the box are labeled as normal force, friction, and weight. The coefficient of static friction between the box and the truck bed must satisfy f ≤ [Symbol]Mg or [Symbol] ≥ v^2/Rg. If the roadway is properly banked, the box will remain in place even if the truck bed is frictionless. The two forces acting on the box in this case are the normal force and the weight, with the normal force being greater in magnitude as it balances gravity. The solution includes the concept of centripetal force=Fn
  • #1
FlyDoc
5
1

Homework Statement



https://word-edit.officeapps.live.com/we/ResReader.ashx?v=00000000-0000-0000-0000-000000000014&n=E1o4139297797.img&rndm=a7b92f15-cc38-471c-8b47-21ae3920eba1&Fi=SDE1B9C4939203BCDE!145&C=5_810_BN1-SKY-WAC-WSHI&ak=m%3Den%2Dus&usid=174cb216-4c95-4a52-a875-f57b5ec0e9fd&build=16.0.3327.1024&waccluster=BN1
1977 B2. A box of mass M, held in place by friction, rides on the flatbed of a truck which is traveling with constant speed v. The truck is on an unbanked circular roadway having radius of curvature R.
a.On the diagram provided above, indicate and clearly label all the force vectors acting on the box.
b.Find what condition must be satisfied by the coefficient of static friction [Symbol] between the box and the truck bed. Express your answer in terms of v, R, and g.
https://word-edit.officeapps.live.com/we/ResReader.ashx?v=00000000-0000-0000-0000-000000000014&n=E1o1230383731.img&rndm=fbbd6a6f-ae47-41de-be84-15d0ce3b1b99&Fi=SDE1B9C4939203BCDE!145&C=5_810_BN1-SKY-WAC-WSHI&ak=m%3Den%2Dus&usid=174cb216-4c95-4a52-a875-f57b5ec0e9fd&build=16.0.3327.1024&waccluster=BN1
If the roadway is properly banked, the box will still remain in place on the truck for the same speed v even when the truck bed is frictionless.

c.On the diagram above indicate and clearly label the two forces acting on the box under these conditions
d.Which, if either, of the two forces acting on the box is greater in magnitude?

Homework Equations



1977B2
https://word-edit.officeapps.live.com/we/ResReader.ashx?v=00000000-0000-0000-0000-000000000014&n=E2o46.img&rndm=344cd4c0-8857-4aca-ba15-160a23c21799&Fi=SDE1B9C4939203BCDE!147&C=5_810_BN1-SKY-WAC-WSHI&ak=m%3Den%2Dus&usid=2c85db80-ac0a-4c4b-9e9b-7f862d69d57e&build=16.0.3327.1024&waccluster=BN1 a.1 = normal force; 2 = friction; 3 = weight

b.Friction, f ≤ [Symbol]N where N = Mg. Friction provides the necessary centripetal force so we have f = Mv2/RMv2/R ≤ [Symbol]Mg, or [Symbol] ≥ v2/Rg
https://word-edit.officeapps.live.com/we/ResReader.ashx?v=00000000-0000-0000-0000-000000000014&n=E2o47.img&rndm=4047adab-2741-441a-bab6-79f3c94d5f4c&Fi=SDE1B9C4939203BCDE!147&C=5_810_BN1-SKY-WAC-WSHI&ak=m%3Den%2Dus&usid=2c85db80-ac0a-4c4b-9e9b-7f862d69d57e&build=16.0.3327.1024&waccluster=BN1
c.
d.from the diagram below, a component of the normal force N[Symbol] balances gravity so N[Symbol] must be greater than mg
https://word-edit.officeapps.live.com/we/ResReader.ashx?v=00000000-0000-0000-0000-000000000014&n=E2o48.img&rndm=09c1b10a-4f51-4049-a012-8cb4dcadee11&Fi=SDE1B9C4939203BCDE!147&C=5_810_BN1-SKY-WAC-WSHI&ak=m%3Den%2Dus&usid=2c85db80-ac0a-4c4b-9e9b-7f862d69d57e&build=16.0.3327.1024&waccluster=BN1

The Attempt at a Solution


I am befuddled by the solution. I can see how Fn is greater than mg by virtue of Fn*cos(theta)=mg, But if I incline the coordinate system, I get Mg*cos(theta)=Fn, making Fn smaller than mg. I would expect Fn to be smaller because the incline is getting closer to vertical, thereby decreasing Fn and leading to slippage. I know this has something to do with Centripetal Force=Fn*sin(theta) but that force is not included on the free body diagram and I can't quite see how that would change the actual normal force. Looking for clarification. Thank you.
 

Attachments

  • AP B 1977B2 (UCM) (1).docx
    70.8 KB · Views: 570
Last edited by a moderator:
Physics news on Phys.org
  • #2
I've attached the question and college board solution
 

Attachments

  • AP B 1977B2 (UCM).docx
    70.8 KB · Views: 919

1. What is UCM and how does it relate to the AP B Exam (1977 B2)?

UCM stands for Uniform Circular Motion and it is a type of motion in which an object moves in a circular path at a constant speed. This concept is often tested in the AP Physics B Exam, specifically in the 1977 B2 question.

2. What is the solution to question #3 in the 1977 B2 AP Physics Exam?

The solution to question #3 involves using the equations for UCM, specifically the centripetal acceleration formula, to calculate the acceleration of an object moving in a circle at a given radius and speed. It also requires understanding the relationship between tangential and centripetal acceleration.

3. Why is question #3 in the 1977 B2 AP Physics Exam confusing?

Question #3 in the 1977 B2 AP Physics Exam can be confusing because it involves multiple concepts and requires a thorough understanding of UCM and its equations. It also requires careful analysis of the given information and applying it correctly to the problem.

4. Are there any tips for solving UCM problems, specifically question #3 in the 1977 B2 AP Physics Exam?

Some tips for solving UCM problems include drawing a diagram to visualize the situation, identifying known and unknown variables, using the appropriate equations, and double-checking your calculations. For question #3 in the 1977 B2 AP Physics Exam, it is also helpful to understand the relationship between tangential and centripetal acceleration.

5. How can I improve my understanding of UCM for the AP Physics Exam (1977 B2)?

To improve your understanding of UCM for the AP Physics Exam, it is important to review the concepts and equations, practice solving problems, and seek help from a teacher or tutor if needed. It can also be helpful to study past exam questions, such as question #3 from the 1977 B2 exam, to become familiar with the types of problems that may be asked.

Back
Top