Physics Help Pls!! Rolling Friction, Lift, Aerodynamic drag!?
Problem 17-7 assumes that the rolling friction is constant at all speeds. This is not exactly true as there is some aerodynamic lift that reduces the apparent weight of the automobile as the speed increases. At 65 km/hr the lift is 300 N and at 120 km/hr it is 900 N. In the light of this further information re-solve Problem 17-7. Assume that rolling friction is proportional to the normal force.
For problem 17-7, the power for 65 km/h = 7.9 kW and for 120 km/hr the power = 32.1 kW. I can do the calculations for this problem based on the steps from the other problem, but I am confused with the lift. The lift will reduce the weight of the car (1500 kg is initial weight given) correct? The rolling friction for 17-7 was 220N. If the weight was not directly used for calculations in the last problem. How do I relate proportionality of lift here?
Solution to 17-7 is here: http://answers.yahoo.com/question/index;_ylt=ArJLyjeDjF93ICA2PJMsFlXsy6IX;_ylv=3?qid=20100108081646AARxq0R
In the original question, it was assumed that rolling friction was constant. With that assumption, we did not need the weight of the vehicle. This new problem changes that and says that we are now going to assume that rolling friction is proportional to the force pushing down on the wheels. Now we need the weight.
At low speeds, the normal force is the same as the weight of the car because there is essentially no lift. The weight is 1500 kg * 9.81 m/s^2 = 14,715 N. We were told that the rolling friction was 220 N at low speed.
At 65 km/h, we are told the lift is 300 N. So we subtract this from the weight of the vehicle to get the downward force on the wheels: Normal force = 14,715 N – 300 N = 14,415 N.
We are told that the rolling friction is proportional to the normal force. So that means the normal force is 220N * (14,415 / 14,715) = 215.5 N. That’s not a whole lot less than before! But now you just use that instead of the 220 N in figuring out the total force opposing the motion of the vehicle (aerodynamic drag plus rolling friction).
At 65 km/h, that’s 220 N + 215.5 N = 435.5 N.
The speed is 65 km/h * 1000 m/km * 1 h/3600 sec = 18.06 m/s
The power will now be P = F * V = 435.5 N * 18.06 m/s = 7863 W = 7.86 kW.
At 120 km/h, we are told the lift is 900 N. So now the downward force is 14715 – 900 = 13815 N.
The rolling friction will now be 220 N 8 (13815/14715) = 206.5 N
As before, the drag goes up in proportion to the speed.
The speed is 120 km/h * 1000/3600 = 33.33 m/s
So drag = 220 N * (33.33/18.06)^2 = 749.8 N
The total force opposing the motion is 749.8 + 206.5 N = 956.4 N
Power = 956.4 N * 33.33 m/s = 31,879 W = 31.9 kW
In the original question, it was assumed that rolling friction was constant. With that assumption, we did not need the weight of the vehicle. This new problem changes that and says that we are now going to assume that rolling friction is proportional to the force pushing down on the wheels. Now we need the weight.
At low speeds, the normal force is the same as the weight of the car because there is essentially no lift. The weight is 1500 kg * 9.81 m/s^2 = 14,715 N. We were told that the rolling friction was 220 N at low speed.
At 65 km/h, we are told the lift is 300 N. So we subtract this from the weight of the vehicle to get the downward force on the wheels: Normal force = 14,715 N – 300 N = 14,415 N.
We are told that the rolling friction is proportional to the normal force. So that means the normal force is 220N * (14,415 / 14,715) = 215.5 N. That’s not a whole lot less than before! But now you just use that instead of the 220 N in figuring out the total force opposing the motion of the vehicle (aerodynamic drag plus rolling friction).
At 65 km/h, that’s 220 N + 215.5 N = 435.5 N.
The speed is 65 km/h * 1000 m/km * 1 h/3600 sec = 18.06 m/s
The power will now be P = F * V = 435.5 N * 18.06 m/s = 7863 W = 7.86 kW.
At 120 km/h, we are told the lift is 900 N. So now the downward force is 14715 – 900 = 13815 N.
The rolling friction will now be 220 N 8 (13815/14715) = 206.5 N
As before, the drag goes up in proportion to the speed.
The speed is 120 km/h * 1000/3600 = 33.33 m/s
So drag = 220 N * (33.33/18.06)^2 = 749.8 N
The total force opposing the motion is 749.8 + 206.5 N = 956.4 N
Power = 956.4 N * 33.33 m/s = 31,879 W = 31.9 kW
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