Question

A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 × 10^3 kg. What is the frictional force on the car?

Answer 1

`f_r = 150.47 N`

Step-by-step explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

`(mv^2)/(r)= N sin \theta + f_r cos \theta`

∑ F_y = 0

`0 = N cos \theta - f_r sin \theta - mg`

`N = (f_rsin \theta + mg)/(cos \theta)`

`(mv^2)/(r)= ((f_rsin \theta + mg)/(cos \theta))sin \theta + f_r cos \theta`

=
`f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta`

`f_r = ((mv^2)/(r)- mg tan\theta)/(sin\theta tan \theta + cos \theta)`

`f_r = ((1.4* 10^3 * 32^2)/(539)- 1.4* 10^(3)* 9.8 * 0.087)/(0.087 * 0.087 + 0.996)`

`f_r = 150.47 N`