A car moving at 37 km/h negotiates a 140 m -radius banked turn designed for 60 km/h . What coefficient of friction is needed to keep the car on the road?
To find the coefficient of friction needed to keep the car on the road, we can use the concept of centripetal force. The car needs sufficient friction between its tires and the road to provide the necessary centripetal force for it to negotiate the banked turn without slipping.
First, we need to determine the speed of the car in terms of the radius of the banked turn. We can use the formula for centripetal force:
F = (mv^2) / r
where F is the centripetal force, m is the mass of the car, v is the speed of the car, and r is the radius of the banked turn.
In this case, we are given the speed of the car (37 km/h) and the radius of the banked turn (140 m). However, we need to convert the speed of the car from km/h to m/s to match the SI unit of meters. To convert km/h to m/s, we divide by 3.6:
v = 37 km/h ÷ 3.6 = 10.28 m/s (rounded to two decimal places)
Now we have the speed of the car (v) and the radius of the banked turn (r), so we can rearrange the formula to solve for the centripetal force:
F = (mv^2) / r
Next, we need to determine the centripetal force required for the car to negotiate the banked turn at a speed of 60 km/h. Again, we convert the speed to m/s:
v = 60 km/h ÷ 3.6 = 16.67 m/s (rounded to two decimal places)
Now we can calculate the centripetal force for the desired speed using the same formula:
F = (mv^2) / r
We know that the mass of the car (m) will cancel out in this calculation because it is the same for both speeds.
Now, we can find the ratio of the actual centripetal force to the required centripetal force:
F_actual / F_required = (mv_actual^2) / (mv_required^2)
F_actual / F_required = (v_actual^2) / (v_required^2)
Substituting the known values:
F_actual / F_required = (10.28^2) / (16.67^2)
F_actual / F_required ≈ 0.394
Lastly, we know that the frictional force between the tires and the road can provide the remaining centripetal force needed to keep the car on the road. The maximum value of the frictional force is given by the coefficient of friction (µ) multiplied by the normal force (N).
F_friction = µN
In this case, the normal force (N) is equal to the weight of the car, which is given by:
N = mg
where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the mass cancels out in the calculation, we can write the equation as:
F_friction = µmg
Rearranging the equation and substituting the derived value for F_friction:
µ = F_friction / mg
µ = (F_actual / F_required) * (v_required^2 / r * g)
Substituting the known values:
µ ≈ 0.394 * ((16.67)^2 / (140 * 9.8))
After performing the calculations, we find that the coefficient of friction needed to keep the car on the road is approximately 0.17 (rounded to two decimal places).