- #1

stunner5000pt

- 1,461

- 2

## Homework Statement

The mass m car is traveling at v on the straight portion of the road and then its speed is uniformly reduced from A to C, which point it comes to rest. Compute the magnitude and direction of the total friction force F exerted by the road on the car (a) just before it passes point B, (b)

just after it passes point B and (c) just before it stops at point C. (d) What is the minimum

coefficient of static friction required between the tires and the road for the car to be able to complete the turn at B?

**2. The attempt at a solution**

Part a - the force of friction as it approaches B

I thought about doing this in two steps:

If we assume that the force of friction is [itex] F_{f} [/itex] then the linear acceleration in the straight part is [itex] a = \frac{F_{f}}{m} [/itex]

so we can use a kinematic formula to determine the velocity at point B

[tex] v_{B}^2 = v^2 + 2a d [/tex]

then

[tex] v_{B}^2 = v^2 - \frac{2 F_{f}}{m} [/tex]

now for the curved section, assuming the road has radius R, the force stays constant, then we can say that the linear acceleration is the same as above. then the angular acceleration is [itex] \alpha = \frac{-F_{f}}{mr} [/itex]

then we can use the rotational version of the above kinematic formula. the velocity at c is zero

[tex] 0 = \omega_{B}^2 + 2 \left( \frac{-F_{f}}{mr} \right) \left( \frac{\pi}{6} \right) [/tex]

we can solve for Vb from the above equation

[tex] v_{B} = \frac{\pi F_{f}}{3mr^3} [/tex]

and then we can substitute the equation just derived into the other expression for vb found above. But is this correct?

What about the curved section? IN the curved section, does the sum of the components of friction (caused by the tangential and normal components) the same as the friction we just calculated? In that case, isn't the answer for b and c the same?

Thanks for your help and input!