# 4 - homework 04 ARMSTRONG, DOMINIC Due: Oct 2 2007, 3:00 am...

This preview shows pages 1–2. Sign up to view the full content.

homework 04 – ARMSTRONG, DOMINIC – Due: Oct 2 2007, 3:00 am 1 Question 1, chap 6, sect 3. part 1 of 5 10 points A car travels at a speed of 29 m / s around a curve of radius 47 m. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 8 Mg μ 29 What is the net centripetal force needed to keep the car from skidding sideways? Correct answer: 50102 . 1 N (tolerance ± 1 %). Explanation: Let : m = 2800 kg , v = 29 m / s , r = 47 m , θ = 29 , and μ = 0 . 631961 Part5 . The centripetal acceleration of the car rounding a curve is a c = v 2 R and the net centripetal force needed to provide such ac- celeration is F c = ma c = mv 2 R = 50102 . 1 N . Question 2, chap 6, sect 3. part 2 of 5 10 points Were there no friction between the car’s tires and the road, what centripetal force could be provided just by the banking of the road? Correct answer: 15210 . 2 N (tolerance ± 1 %). Explanation: In the absence of friction, there are only two forces acting on the car, namely its weight mVg and the normal force V N provided by the road. The weight is directed vertically down while the normal force is directed perpendicular to the banked road surface and thus at the angle θ from the vertical. The centripetal force due to banking comes from the horizontal component of the normal force, F banking c = N sin θ . The free body diagram in the vertical di-

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/07/2008 for the course PHY 301 taught by Professor Swinney during the Fall '07 term at University of Texas at Austin.

### Page1 / 4

4 - homework 04 ARMSTRONG, DOMINIC Due: Oct 2 2007, 3:00 am...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online