- #1

thebrandon

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## Homework Statement

A curve with a 125 radius on a level road is banked at the correct angle for a speed of 20. If an automobile rounds this curve at 30 , what is the minimum coefficient of static friction between tires and road needed to prevent skidding?

## Homework Equations

F=ma

centripetal acceleration = (v^2)/r

## The Attempt at a Solution

My teacher likes us to leave the x and y axes in there standard places, and create new axes in the parallel and perpendicular directions. In this case, the parallel axis would be the track, and the perpendicular axis would be perpendicular to the track.

I separated all forces into their parallel and perpendicular components:

Parallel gravity: mgsin(theta)

Perpendicular gravity: mgcos(theta)

Normal force (in the perpendicular direction): mgcos(theta)

I then found the necessary inward force to maintain a circle:

Inward force: (v^2)/R * m = 400/125 * m = 3.2m

Separated into parallel and perpendicular components:

Parallel: 3.2mcos(theta)

Perpendicular: 3.2msin(theta)

If the car is not sliding and there is no friction, then the parallel gravity force is equal to the parallel component of this necessary inward force: 3.2mcos(theta)=mgsin(theta) this leads to theta=18 degrees

Now if a car is moving 30m/s, then the new necessary inward force is 900/125 * m = 7.2m

The components of this are: parallel: 7.2mcos18 , Perpendicular: 7.2 sin18

The friction can be denoted as f=u*normal force=u*mgcos18

If the car is not sliding then 7.2mcos18= mgsin18+ umgcos18

This solves out to u = .41

The answer is supposed to be .33

Where did I go wrong?

Second problem:

## Homework Statement

The 4.00- block in the figure is attached to a vertical rod by means of two strings. When the system rotates about the axis of the rod, the strings are extended as shown in the diagram and the tension in the upper string is 79.0 .

A) What is the tension in the lower cord?

B)How many revolutions per minute does the system make?

C)Find the number of revolutions per minute at which the lower cord just goes slack.

## Homework Equations

F=ma

Centripetal acceleration = (v^2)/R

## The Attempt at a Solution

(1^2) + (L^2) = (1.25^2) , solving this gives me that the mass is .75 m from the rod

Arctan(1/.72) = theta , theta = 53.13

Necessary inward force: (v^2)/R * m

For upper string: ((v^2)/.75) *m = 79cos 53.13 this leads to m*(v^2) =35.55

That was all I could figure out so far.

Any help would be greatly appreciated. I have been working on this problems for a day and a half, and I just can't seem to solve them. Thanks in advance.