A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coefficient of static friction between the car's tires and the road is 0.300. What is the magnitude of the force exerted by friction on the car?

Question

Cade Mayer

Physics

13 June, 00:55

A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coefficient of static friction between the car's tires and the road is 0.300. What is the magnitude of the force exerted by friction on the car?

+5

Answers (2)

Know the Answer?

Bradshaw · Answer 1

The magnitude of force is 1593.4N

Explanation:

The sum of the horizontal components of the friction and the normal force will be equal to the centripetal force on the car. This can be represented as

fcostheta + Nsintheta = mv^2/r

Where F = force of friction

Theta = angle of banking

N = normal force

m = mass of car

v = velocity of car

r = radius of curve

The car has no motion in the vertical direction so the sum of forces = 0

The vertical component of the normal force acts upwards whereas the weight of the car and the vertical component friction acts downwards.

Taking the upward direction to be positive, rewrite the equation above to get:

Ncos thetha = mg - fsintheta = 0

Ncistheta = mg + fain theta

N = mg/cos theta + sintheta / costheta

fcostheta + [mg/costheta + ftan theta] sin theta = mv^2/r

Substituting gives:

f = (1 / (costheta + tanthetasintheta) + mgtantheta = mv^2/r - mgtantheta)

Substituting given values into the above equation

f = 1 / (cos25 + tan 25) (sin25) [ 600*30/120 - (600*9.81) tan

f = 1593.4N25

Valentin Rodriguez · Answer 2

1593.4 N

Explanation:

The road is banked at an angle of 25°. The frictional force along the banked road f. The horizontal component of the car's weight along the road is F = mgsinθ. The horizontal net force along the road is thus F₁ = F - f. F₁ also equals the centripetal force on the car as it goes round the curve. So, F₁ = mv²cosθ/r

F - f = mv²cosθ/r

mgsinθ - f = mv²cosθ/r

f = mgsinθ - mv²cosθ/r

f = m (gsinθ - v²cosθ/r)

Given m = mass of car = 600 kg, g = 9.8 m/s², θ = banking angle = 25°, v = speed of car = 30 m/s and r = radius of curved road = 120 m

f = 600 (9.8sin25 - 30²cos25/120)

f = 600 (4.1417 - 6.7973)

f = 600 (-2.6556)

f = - 1593.4 N

This is the friction between the road and the tyres. The magnitude of the frictional force exerted on the car is 1593.4 N

Since f = μmgcosθ where μ = coefficient of static friction = 0.3

f = 0.3 * 600 * 9.8cos25 = 1598.73 N