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

AI Thread Summary
The discussion revolves around the physics problem from the 1977 AP B Exam regarding a box on a truck traveling in a circular path. Participants express confusion about the forces acting on the box, particularly the normal force and its relationship to gravitational force when the truck is on a banked curve. The key point is that the normal force must be greater than the weight of the box to prevent slippage, which is counterintuitive for some when considering the incline's effect on force components. Clarification is sought on how centripetal force interacts with the normal force and the implications for the box's stability. Understanding these dynamics is crucial for solving the problem accurately.
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Homework Statement



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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.
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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
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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
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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.
 

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I've attached the question and college board solution
 

Attachments

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